Answer:
10s
Explanation:
Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.
If acceleration is constant, then the velocity is changing by a constant amount.
With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:
[tex]v=at+v_o[/tex] where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.
[tex]v=at+v_o[/tex]
[tex](1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})[/tex]
[tex]1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t[/tex]
[tex]\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}[/tex]
[tex]10\text{ [s]}=t[/tex]
So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.
Verification:
In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.
Recognizing that the acceleration is [tex]a=\dfrac{100 [\frac{m}{s}]}{1[s]}[/tex], the velocity increases by 100 units [m/s] every second.
At time 0[s], the velocity is 0[m/s]
At time 1[s], the velocity is 100[m/s]
At time 2[s], the velocity is 200[m/s]
At time 3[s], the velocity is 300[m/s]
At time 4[s], the velocity is 400[m/s]
At time 5[s], the velocity is 500[m/s]
At time 6[s], the velocity is 600[m/s]
At time 7[s], the velocity is 700[m/s]
At time 8[s], the velocity is 800[m/s]
At time 9[s], the velocity is 900[m/s]
At time 10[s], the velocity is 1000[m/s]
So, indeed, after 10 seconds, the velocity reaches 1000 m/s
Select the correct answer. Which segment of this graph shows a decreasing velocity? Graph representing Time in seconds on the x-axis and position in miles on the y-axis. The line starts at point A (0, 20), and passes points B (20, 20), C (40, 40), D (60, 60), E (80, 60), and F (100, 80) A. A to B B. B to C C. C to D D. D to E E. E to F
The portion that shows a decreasing velocity is D to E.
What is a position - time graph?The position time graph is used to describe the motion of a body. We know that the slope of a velocity time graph is the velocity of the graph.
If we look at the graph as shown in the image attached to the answer, we can see that the portion that shows a decreasing velocity is D to E.
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Answer:
D - E
Explanation:
hope this help
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 96.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 27.0 kg.
(a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (Select all that apply.)
the wagon
the child in the wagon
the children outside the wagon
(b) Draw a free body diagram, including the weight and all other forces acting on the system. (Do this on paper. Your instructor may ask you to turn in this diagram.)
(c) Calculate the acceleration.
m/s2
(d) What would the acceleration be if friction is 21.0 N?
(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.
(b) The acceleration of the child-wagon system is 0.33 m/s².
(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.
Net force on the third child
Apply Newton's second law of motion;
∑F = ma
where;
∑F is net forcem is mass of the third childa is acceleration of the third child∑F = 96 N - 75 N - 12 N = 9 N
Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;
the wagon the children outside the wagonFree body diagram→ → Ф ←
1st child friction wagon 2nd child
Acceleration of the child and wagon systema = ∑F/m
a = 9 N / 27 kg
a = 0.33 m/s²
When the frictional force is 21 N∑F = 96 N - 75 N - 21 N = 0 N
a = ∑F/m
a = 0/27 kg
a = 0 m/s²
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A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0 m. (It is not a very smart idea to play soccer on the roof of tall buildings.)
The ball is kicked with a speed of v0 = 15.10 m/s at an angle of θ = 74.1° with respect to the horizontal. The mass of a size 5 soccer ball is m = 450 g. When the ball lands on the other building, its speed is 19.89 m/s.
How much energy was lost to air friction? The ball is kicked without a spin.
The energy lost to the air due to friction is 37.7 J.
Energy lost to the air due to frictionThe energy lost to the air due to friction is determined from the change in kinetic energy of the ball.
ΔK.E = K.Ef - K.Ei
ΔK.E = ¹/₂mv² - ¹/₂mu²
ΔK.E = ¹/₂m(v² - u²)
ΔK.E = ¹/₂(0.45)(19.89² - 15.1²)
ΔK.E = 37.7 J
Thus, the energy lost to the air due to friction is 37.7 J.
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how mesuremerment of gravity using simple pendulum
Answer:
See below
Explanation:
Set up your pendulum
measure its length and time the period ( you could time 100 of them and divide the time result by 100 to get the period, T)
then use
T = 2 pi sqrt (L/g) T = period L = length g = gravity
You cover the following displacements every day going to school: d1=50 m, E and d2=95 m, N. You do this for 12 minutes. a) What is your speed in m/s? b) What is your velocity in m/s?
The speed will be 0.2 m/s and the velocity will be 0 m/s.
Speed = Total Distance / Total time
We have given total distance as ( 50 + 95 ) metres and total time as 12 minutes or we can say 720 seconds.
Speed = 145/ 720 m/s
Speed = 0.2 m/s
Velocity = Total Displacement / Total time
As the initial and final is the home, hence the net displacement is 0 in that case.
In this case also the total time we have given is 12 minutes or we can say 720 seconds.
Velocity = 0 / 720 m/s
Velocity = 0 m/s
So to conclude with we can say that the speed is 0.2 m/s and the velocity is 0 m/s.
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3. The following sound waves have what velocity? (convert cm into m in your calculations.)
a. 20.0 Hz, λ of 17.2 m.
b. 200.0 Hz, λ of 1.72 m.
c. 2,000.0 Hz, λ of 17.2 cm.
d. 20,000.0 Hz, λ of 1.72 cm.
The sound waves with the frequency 20 Hz, 200 Hz, 2,000 Hz, and 20,000 Hz have a velocity of 344 m/s.
What are sound waves?A sound wave is a sample resulting from the motion of strength traveling thru a medium (inclusive of air, water, or every other liquid or stable matter) because it propagates from the supply of the sound. Sound waves are created and convey strain waves, for example, a cellphone.
The frequency can be given as:
Frequency = Velocity/ Wavelength
Velocity = Frequency * Wavelength
The velocity of the following sound waves is given as:
Frequency = 20 HzWavelength = 17.2 m
Velocity = 20 * 17.2 m/s
Velocity = 344 m/s
Frequency = 200 HzWavelength = 1.72 m
Velocity = 200 * 1.72 m
Velocity = 344 m/s
Frequency = 2,000 HzWavelength = 17.2 cm = 0.172 m
Velocity = 2,000 * 0.172 m/s
Velocity = 344 m/s
Frequency = 20,000 HzWavelength = 1.72 cm = 0.0172 m
Velocity = 20,000 * 0.0172 m/s
Velocity = 344 m/s
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2 red and 2 blue overlapping balls in the center are surrounded by a green, fuzzy, circular cloud with a white line running through it. 2 green balls sit on the white line, and a line leading a bracket around the balls is labeled A. A line leading to a bracket overlapping the white line is labeled B. Which is a characteristic of the part of the atom marked "B”? It is very dense. It has a large mass. It is negatively charged. It is where neutrons move.
A characteristic of the part of the atom marked "B” is that it has a large mass. Option B. This is further explained below.
What is atom?Generally, Atom, the lowest possible unit into which matter may be split without releasing electrically charged particles.
In conclusion, the protons and neutrons are represented by the red and white spheres, respectively. Since it indicates both protons and neutrons, the line preceding the bracket around the balls (designated A) stands in for the atom's nucleus.The mass of the "B" section of the atom is very high.
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Answer:
it's b
Explanation:
i did the ed quiz
Please help me with this answer
The position of the car at the given time is 18 m.
Position of the car at t = 6.0 s
The position of the car at the given time is calculated as follows;
x = vt
where;
v is the velocity of the car; from the graph v = 3 m/st is time of motion = 6.0 sx = 3 m/s x 6.0 s
x = 18 m
Thus, the position of the car at the given time is 18 m.
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A roller coaster has a mass of 450 kg. It sits at the top of a hill with height 49
m. If it drops from this hill, how fast is it going when it reaches the bottom?
Answer:
Approximately [tex]31\; {\rm m\cdot s^{-1}}[/tex] assuming that there is no friction on the rollercoaster, and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
Make use of the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where:
[tex]v[/tex] is the final velocity of the moving object,[tex]u[/tex] is the initial velocity of the object,[tex]a[/tex] is the acceleration of the object, and[tex]x[/tex] is the displacement of the object.If there is no friction on the rollercoaster, the acceleration of the rollercoaster would be equal in magnitude to the gravitational field strength, [tex]g[/tex]:
[tex]a = g = 9.81\; {\rm m \cdot s^{-2}}[/tex].
The initial velocity of this rollercoaster is [tex]0\; {\rm m\cdot s^{-1}}[/tex] (that is, [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]) since the rollercoaster was initially stationary. The displacement of this rollercoaster would be [tex]x = 49\; {\rm m}[/tex] (same as the height of the hill.)
Rearrange the equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] to find an expression for the final velocity [tex]v[/tex] of this rollercoaster:
[tex]\begin{aligned}v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 9.81\; {\rm m\cdot s^{-2} \times 49\; {\rm m} + (0\; \rm m\cdot s^{-1}})^{2}}\\ &\approx 31\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
I need a super tragic past for my desired reality, the reality I'm going to is Naruto.
Thanks
To create a tragic past for your character, you would have to write a plot that shows a sad ending to a good story.
What is a Tragic Story?This refers to a type of story that has negative emotions such as sadness, death, etc.
Hence, we can see that for example, you could write about two lovers and how despite all the struggles they faced, eventually died in mysterious circumstances.
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A cube, whose mass is 0.660 kg, is attached to a spring with a force constant of 106 N/m. The cube rests upon a frictionless, horizontal surface (shown in the figure below).
The cube is pulled to the right a distance A = 0.140 m from its equilibrium position (the vertical dashed line) and held motionless. The cube is then released from rest.
(a) At the instant of release, what is the magnitude of the spring force (in N) acting upon the cube?
________N
(b) At that very instant, what is the magnitude of the cube's acceleration (in m/s2)?
_____ m/s2
(c)
In what direction does the acceleration vector point at the instant of release?
- The direction is not defined (i.e., the acceleration is zero).
-Away from the equilibrium position (i.e., to the right in the figure).
- Toward the equilibrium position (i.e., to the left in the figure).
- You cannot tell without more information.
***Please refer to attached photo
Answer:
F = - F x force acting on cube
F = -106 N/n * .14 m = -14.8 N direction opposite to force
a = F / m = -14.8 N / .66 kg = -22.5 m/s^2 acceleration of cube
direction is towards equilibrium point (opposite F)
Hello!
I have an equation where I can't really find To (Initial temperature). The equation is from the formula Newton's law of cooling. To is the initial temperature of the object and ln is the natural log. The -0.00124 is the constant k multiplied by the time.
I just need to find the initial temp.
Answer:
[tex]T_0=80695.17162...[/tex]
Explanation:
Given equation:
[tex]\ln \left(\dfrac{T_0-100}{T_0}\right)=-0.00124[/tex]
To solve the given equation:
[tex]\textsf{Apply log rules}: \quad e^{\ln (x)}=x[/tex]
[tex]\implies \dfrac{T_0-100}{T_0}=e^{-0.00124}[/tex]
Multiply both sides by T₀:
[tex]\implies T_0-100=T_0e^{-0.00124}[/tex]
Add 100 to both sides:
[tex]\implies T_0=T_0e^{-0.00124}+100[/tex]
Subtract [tex]T_0e^{-0.00124}[/tex] from both sides:
[tex]\implies T_0-T_0e^{-0.00124}=100[/tex]
Factor out the common term T₀:
[tex]\implies T_0(1-e^{-0.00124})=100[/tex]
Divide both sides by [tex](1-e^{-0.00124})[/tex]
[tex]\implies T_0=\dfrac{100}{1-e^{-0.00124}}[/tex]
Carry out the calculation:
[tex]\implies T_0=\dfrac{100}{1-0.99876...}[/tex]
[tex]\implies T_0=\dfrac{100}{0.001239231...}[/tex]
[tex]\implies T_0=80695.17162...[/tex]
What is my speed for this?
A single frictionless roller-coaster car of mass m = 750 kg tops the first hill with speed v= 15 m/s at height h = 40 m as shown
The speed of cars at A and B will be 25 m/sec.
What is speed?Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its si unit is m/sec.
Given data;
Mass of roller coaster,m =750 kg
Height at location 1,h₁ = 40 m
Height at location 2,h₂ = 302 m
The velocity of car A,[tex]\rm v_A[/tex]
The velocity of the car B, [tex]\rm v_B[/tex]
The velocity of the car A and B is the same is found as;
[tex]\rm v_A = v_B = v = \sqrt{v_0^2 + 2g(h_1-h_2} )\\\\ v = \sqrt{15^2 + 2 \times 10 (40-30)} \\\\ v = \sqrt{225+400} \\\\ v = 25 \ m/sec[/tex]
Hence the speed of cars at A and B will be 25 m/sec
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A ball slows down as it rolls up a hill. What is the value of the change in its kinetic energy, ΔEk?
Positive
Zero
Negative
For a ball that slows down as it rolls up a hill the value of the change in its kinetic energy, ΔEk is: Negative
Meaning of Kinetic energyKinetic energy can be defined as the ability assumed by an object in motion.
Kinetic energy is also said to be a form of mechanical energy that is possessed by an object to be in motion.
Normally kinetic energy is believed to be positive by when it begins to slow down a negative effect is experienced
In conclusion, For a ball that starts rolling down a ramp the value of the change in its kinetic energy, ΔEk is: Positive
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The cannon on a battleship can fire a shell a maximum distance of 36.0 km.
(a) Calculate the initial velocity of the shell.
m/s
(b) What maximum height does it reach? (At its highest, the shell is above a substantial part of the atmosphere--but air resistance is not really negligible as assumed to make this problem easier.)
m
(c) The ocean is not flat, since the earth is curved. How many meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface at the ship?
mDoes your answer imply that error introduced by the assumption of a flat earth in projectile motion is significant here? (Select all that apply.)
The error is insignificant compared to the distance of travel.
The error is significant compared to the distance of travel.
The error could be significant compared to the size of a target.
The error is insignificant compared to the size of a target.
(a)The initial velocity of the shell will be 594.27 m/sec
(b)The maximum height it reaches will be 9000 m.
c)101.249 m meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface of the ship.
d)The error could be significant compared to the size of a target. Option C is correct.
What is velocity?The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
m is the mass of the block = Kg.
u is the initial velocity of fall = m/sec
h is the distance of fall = m
g is the acceleration of free fall = m/sec²
v is the hitting velocity of =?
a)
The range of the projectile is;
[tex]\rm R = \frac{u^2 sin 2 \theta }{g} \\\\ 36 \times 10^ 3 = \frac{U^2 sin 45^0}{9.81} \\\\ U= 594.27 \ m/sec[/tex]
b)
The maximum height of the projectile is;
[tex]\rm H = \frac{u^2 sin 2 \theta }{2g} \\\\ H = \frac{(594.27)^2\times (sin 45)^2}{2 \times 9.81 } \\\\ H = 9000 \ m[/tex]
c)
The distance between its surface and the ship, measured in a horizontal arc parallel to the surface, will be 36.0 kilometers. The distance from the lower surface is found as;
[tex]\rm( R_e + h)^2 = R_e^2+(36)^2 \\\\ (R_e)^2 = h^2+2R_e h= R_e^2 + 12196 \\\\ h^2 + 12800 h - 1296 = 0 \\\\ h = 101.249 \ m[/tex]
d)
An error is a mistaken or erroneous action. In some contexts, an error is interchangeable with a mistake.
The difference between the calculated value and the original value is known as the error. The inaccuracy may be large in comparison to the target's size. Option C is correct.
Hence the initial velocity of the shell, maximum height, and the distance from the lower surface will be 594.27 m/sec,9000 m, and 101.249 m and option c for question d are correct respectively.
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A 21.0 kg uniform beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 66.0° what is the tension in the cable?
What is the vertical component of the force exerted by the hi.nge on the beam?
The tension in the cable is 112.64 N and the vertical component of the force exerted by the hi.nge on the beam is 93.16 N.
Tension in the cableApply the principle of moment and calculate the tension in the cable;
Clockwise torque = TL sinθ
Anticlockwise torque = ¹/₂WL
TL sinθ = ¹/₂WL
T sinθ = ¹/₂W
T = (W)/(2 sinθ)
T = (21 x 9.8)/(2 x sin66)
T = 112.64 N
Vertical component of the forceT + F = W
F = W - T
F = (9.8 x 21) - 112.64
F = 93.16 N
Thus, the tension in the cable is 112.64 N and the vertical component of the force exerted by the hi.nge on the beam is 93.16 N.
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The mass of a star is 1.210×10[tex]^{31}[/tex] kg and it performs one rotation in 20.30 days. Find its new period (in days) if the diameter suddenly shrinks to 0.350 times its present size. Assume a uniform mass distribution before and after.
Since the diameter of the star suddenly shrinks to 0.350 times its present size, the new period of the star is 4.2 days
To find the new period, we need to know Kepler's third law of planetary motion
What is Kepler's third law of planetary motion?Kepler's third law of planetary motion states that the square of the period of orbit of a planet, T equals the cube of its radius, R.
So, T² ∝ R³
and T₂²/T₁² = R₂³/R₁³ where
T₁ = initial period of star = 20.30 days, T₂ = final period of star after shrinking, R₁ = initial radius of star and R₂ = final radius of star after shrinkingGiven that the diameter suddenly shrinks to 0.350 times its present size and its radius is proportional to its diameter. So, its radius shrinks 0.350 times its initial size. Also, since there is a uniform mass distribution before and after, the mass remains the same.
So, R₂ = 0.350R₁
The new period of the starMaking T₂ subject of the formula, we have
T₂ = [√(R₂/R₁)³]T₁
Substituting the values of the variables into the equation, we have
T₂ = [√(R₂/R₁)³]T₁
T₂ = [√(0.350R₁/R₁)³]20.30 days
T₂ = [√(0.350)³]20.30 days
T₂ = [√0.042875]20.30 days
T₂ = [0.2071]20.30 days
T₂ = 4.2 days
So, the new period of the star is 4.2 days
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The water is flowing through the horizontal constricted pipe. The pressure at one end is 4500Pa, speed is 3m/s and area of cross-section is A, Calculate the speed and pressure at another end where the area of cross-section is A/3.
The speed and pressure at another end will be 9 m/sec and -31500 pascals.
What is gauge pressure?The difference between absolute pressure and atmospheric pressure is known as gauge pressure. Relative pressure is another name for gauge pressure.
Given data in problem is;
For horizontal pipe, Z₁=Z₂
Pressure at end 1, P₁ = 4500Pa
Speed at end 1, V₁ = 3 m/sec
Speed at end 2, V₂ = ? m/sec
Area of cross-section at end 1,A₁ =A
Area of cross-section at end 1,A₂ = A/3
From the continuity equation;
A₁V₁=A₂V₂
A× 3 = (A/3)×V₂
V₂ = 9 m/sec
From Bernoulli's equation;
[tex]\rm \frac{P_1}{\rho g} + \frac{v_1^2}{2g} +Z_1 = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} +Z_2 \\\\ \frac{P_1-P_2}{\rho g } = \frac{v_2^2-v_1^2}{2g} \\\\ \frac{P_1-P_2}{\rho} = \frac{v_2^2-v_1^2}{2} \\\\ \frac{4500-P_2}{1000} = \frac{9^2-3^2}{2} \\\\ 4500 -P_2 = 36000 \\\\ P_2 = - \ 31,500 \ pa[/tex]
Hence the speed and pressure at another end will be 9 m/sec and -31500 pascals.
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The mass of a star is 1.210×1031 kg and it performs one rotation in 20.30 days. Find its new period (in days) if the diameter suddenly shrinks to 0.350 times its present size. Assume a uniform mass distribution before and after.
The new period will be 2.486 days.
What is the period?
The period is found as the ratio of the angular displacement and the angular velocity. Its unit is the second and is denoted by t. The value of time needed to complete the rotation is the total period.
Given data;
Mass of a star,m= 1.210×10³¹ kg
The time period for one rotation of the star, T = 20.30 days
D' = 0.350 D
R' = 0.350 R
From the law of conservation of angular momentum;
[tex]\rm I \omega = I' \omega' \\\\ \frac{2}{5} MR^2 \times \frac{2 \pi }{T}=\frac{2}{5} MR'^2 \\\\ R^2 \times \frac{1}{T}= R'^2 \times \frac{1}{T} \\\\ T' = \frac{R'^2T}{R^2} \\\\ T' = \frac{(0.350 R)^2 \times 26.1 }{R^2} \\\\T' = 0.1225 \times 20.30 \\\\ T'= 2.486 \ days[/tex]
Hence, the new period will be 2.486 days.
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A sphere of radius r has surface area
A = 4r2and volume V = (4/3r^3) The radius of sphere 2 is triple the radius of sphere 1.
(a)
What is the ratio of the areas, A2/A1?
(b)
What is the ratio of the volumes, V2/V1?
Answer:
A2/A1 = 9
V2/V1 = 27
Explanation:
Area of Sphere 1; A1 = 4πr²
Volume of Sphere 1; V1 = (4/3)πr³
Area of Sphere 2; A2 = 4π(3r)² = 9*4πr²
Volume of Sphere 2; V2 = (4/3)π(3r)³ = 27*(4/3)πr³
Thus;
A2/A1 = 9*4πr²/4πr² = 9
V2/V1 = [27*(4/3)πr³]/[(4/3)πr³] = 27
Complete the sentence to describe how light behaves as it relates to its nature (waves or particles). light travels as and interacts with matter as .
Light behaves as it relates to its nature (waves or particles). light travels as and interacts with matter as wave particles.
What is light?Electromagnetic radiation that is within the region of the electromagnetic spectrum that the human eye can see is referred to as light or visible light.
The main characteristics of light include intensity, polarization, frequency or wavelength spectrum, and propagation direction. One of the fundamental constants of nature is its speed in a vacuum, which is 299 792 458 meters per second (m/s).
Light can be categorized according to two theories: the first categorizes light as particles, and the second categorizes light as waves.
The main difference between the wave and particle natures of light is that the former asserts that light can act as an electromagnetic wave while the latter asserts that light is made up of tiny particles known as photons.
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Answer: 1.waves 2.particles
Explanation: just did it!<3
An extended body (not shown in the figure) has its center of mass at the origin of the reference frame. In the case below give the direction for the torque τ with respect to the center of mass of the body due to force F acting on the body at a location indicated by the vector r.
Options are:
X
-Y
Z
-Z
The direction of torque τ this method is mathematically given as
D=X
Option A is correct.
What is the direction of torque?Generally, the equation for torque is mathematically given as
τ = r X F
Hence to decipher the torque direction with respect to the center of mass of the body due to force F acting on the body at a location indicated by the vector r
We utilize our right hand.Place our right-hand fingers along the path of rPlace our right-hand palm on F Then slowly we sweep r into F.The path or direction of the thumb will provide the direction of the torque.In conclusion, the direction of this method is
D=X Option A.
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The farther a star is away from Earth, the more it is
Answer:
• black hole - one of the most dense objects in the universe
• red giant - a very large star
• white dwarf - an old, very dense hot star that is cooling
• nebula - a mass of gas and dust
An extended body (not shown in the figure) has its center of mass at the origin of the reference frame. In the case below give the direction for the torque τ with respect to the center of mass of the body due to force F acting on the body at a location indicated by the vector r.
Options are:
-Y
x
Z
-Z
+ X axis.
The equation for torque is:
[tex]\tau = r \times F[/tex]
Torque is the CROSS-PRODUCT of 'r' and 'F'.
We can use the right-hand rule to determine the direction of torque.
Looking at the diagram, with an open hand, point your fingers towards the top of the picture (Along the +Z axis in the direction of the 'r' vector.)
Then, curl your fingers in the direction of the force vector, (along the -Y axis).
Your thumb will stick out in the direction of the torque, or the +X axis.
You can also just eliminate any options that involve axes that have already been used for 'r' and 'F'. In this case, there are components in the Z and Y axis, so this leaves the 'X' axis to be considered.
A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg.
If the tension in rope A is 7.42x10^2 N and the tension rope B carries is 7.52x10^2 N, what is the tension in rope C?
The tension in rope C is determined as 122.23 N.
Tension in rope CThe tension in rope C is calculated as follows;
B² = A² + C²
C² = B² - A²
C² = (752²) - (742²)
C² = 14,940
C = √14,940
C = 122.23 N
Thus, the tension in rope C is determined as 122.23 N.
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The stool is 120 cm tall. How tall is it in inches? Note: there are 2.54 cm in 1 inch
Answer:
47.24 in
Explanation:
120 / 2.54 = 47.24
Pls help !!
What must be the same for two resistors that are connected in parallel?
OA. The resistance of each
OB. The voltage drop across each
OC. The power used by each
OD. The current in each
Answer
its c
Explanation:
When resistors are connected in parallel combination the voltage drop across each resistor is same. However , the total current gets branched into different amount .
Question :
— What must be the same for two resistors that are connected in parallel ?
AnswEr :
When the two resistors are connected in parallel therefore , the Voltage drop across each resistors will be same .Option B seems to be correct answer .
Problem 3: A block of mass m= 15.5 kg rests on an inclined plane with a coefficient of
static friction of us = 0.11 between the block and the plane. The inclined plane is L = 6.8 m long and
it has a height of h = 3.05 m at its tallest point.
A. What angle in degrees does the plane make with respect to the horizontal? B. What is the magnitude of the normal force in newtons that acts on the block? C. What is the component of the force of gravity along the plane, fgx in newtons? D. Write an expression in terms of degrees, the mass, the coefficient of the static friction, and the gravitational constant, for the magnitude of the force due to static friction, just before the block begins to slide.
(a) The angle the plane makes with respect to the horizontal is 27⁰.
(b) The magnitude of the normal force acting on the block is 135.3 N.
(c) The component of the force of gravity along the plane, fgx is 68.96 N.
(d) The expression for the net force on the block is mgsinθ - μmgcosθ = 0.
Angle of the plane with respect to the horizontalsin θ = h/L
where;
θ is the angle of inclinationh is height of the inclineL is length of the planesin θ = 3.05/6.8
sin θ = 0.4485
θ = arc tan(0.4485)
θ = 26.7⁰
θ ≈ 27⁰
Normal force on the blockFₙ = mgcosθ
Fₙ = (15.5)(9.8)cos(27)
Fₙ = 135.3 N
Parallel component of the force on the planeFx = mgsinθ
Fx = (15.5)(9.8) sin(27)
Fx = 68.96 N
Net force on the block∑F = 0
Fx - Ff = 0
mgsinθ - μmgcosθ = 0
Thus, we can conclude the following;
the angle the plane makes with respect to the horizontal is 27⁰, the magnitude of the normal force acting on the block is 135.3 N,the component of the force of gravity along the plane, fgx is 68.96 N, and the expression for the net force on the block is mgsinθ - μmgcosθ = 0.Learn more about static friction here: https://brainly.com/question/13680415
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Using appropriate technology, you determine that Earth's magnetic field at your location has a magnitude of 5.88 10-5 T and is directed north at an angle of 56° below the horizontal. You charge a metallic projectile and fire it north with a speed of 663 m/s and at an angle of 14° above the horizontal. Determine the sign (+ or −) and charge of the projectile, if the magnetic force on the projectile is 3.22 10-10 N, directed due west.
C
The sign (+ or −) and charge of the projectile, if the magnetic force on the projectile is 3.22 10-10 N, directed due west is mathematically given as
q=+1.11524613*10^{−4}C
What is the sign (+ or −) and charge?Generally, the equation for Force is mathematically given as
[tex]F=Bqvsin\theta[/tex]
Therefore
[tex]3.22 10^{-10}= 5.88 10^{-5 } *633*sin 56*q\\\\q=\frac{3.22 10^{-10}}{5.88 10^{-5 } *633*sin 56}[/tex]
q=+1.11524613*10^{−4}C
In conclusion, the sign (+ or −) and charge
q=+1.11524613*10^{−4}C
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