The angle θ made by the rope with the box is equal to 25.17 degrees.
If the rope is pulled at an angle A and the tension in the rope was T, then the rope will have two components of forces,
One Vertical which is TsinA and one along horizontal which is TcosA,
And we know,
W = F.s.cosα
Where,
F is the force applied,
s is the Displacement along force,
α is the angle between them,
We know, the box is moving forward, so the angle between the vertical component of tension and displacement is 90 degrees, so the vertical component of tension will not do any work, and the angle between displacement and the horizontal component is 0 degrees,
So, the work done by the horizontal component is 0 degrees,
So, we cam write,
W = TcosA.D
where D is the displacement of the box, which is given to be 23m,
T is the tension in the string which is 120N,
W is the work done by the rope on the box which is 25000J,
so, putting all the values,
2500 = 120.cosA.23
cosA = 2500/120x23
cos A = 0.9
So, A is 25.17 degrees.
The angle made by the rope is 25.17 degrees,
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Calculate the wavelength of a light that has an energy of 9.55x10^-19J.
ANSWER
[tex]2.08\cdot10^{-7}m[/tex]EXPLANATION
To calculate the wavelength of light, we have to apply the formula for the energy of light:
[tex]E=\frac{hc}{\lambda}[/tex]h = Planck's constant
c = speed of light
λ = wavelength
Therefore, we have that:
[tex]\lambda=\frac{hc}{E}[/tex]The wavelength of the light is:
[tex]\begin{gathered} \lambda=\frac{6.626\cdot10^{-34}\cdot3\cdot10^8}{9.55\cdot10^{-19}} \\ \lambda=2.08\cdot10^{-7}m \end{gathered}[/tex]number 16part b: what is the speed of the combined vehicles after the collison?
velocity = p/m
p = momentum= 51,000 kgm/s
m = mass = 3,000+2,000= 5,000kg
v= 51,000 kgm/s / 5000 kg = 10.2 m/s
16. On the moon, Bob weighs 160 N while on earth Fred weighs 882 N Who has thegreater mass?
We are given the weight of two people, one on the moon and the other on earth. To determine the mass we will use the following formula:
[tex]W=mg[/tex]Where:
[tex]\begin{gathered} W=\text{ weight,}\lbrack N\rbrack \\ m=\text{ mass,}\lbrack kg\rbrack \\ g=\text{ acceleration of gravity, }\lbrack\frac{m}{s^2}\rbrack \end{gathered}[/tex]Now, we solve for the mass "m" by dividing both sides by "g":
[tex]\frac{W}{g}=m[/tex]Now, for the case of the moon we have that the acceleration of gravity is:
[tex]g_{moon}=1.62\frac{m}{s^2}[/tex]Plugging in the values:
[tex]\frac{160N}{1.62\frac{m}{s^2}}=m[/tex]Solving the operations:
[tex]98.77kg=m[/tex]Now, for the case of the earth we have:
[tex]g_{\text{earth}}=9.8\frac{m}{s^2}[/tex]Plugging in the values:
[tex]\frac{882N}{9.8\frac{m}{s^2}}=m[/tex]Solving the operations:
[tex]90kg=m[/tex]Therefore, the greater mass is the mass of Bob.
A truck with mass m has a brake failure while going down an icy mountain road of constant downward slope angle α (Figure 1). Initially the truck is moving downhill at speed v0. After careening downhill a distance L with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle β. The truck ramp has a soft sand surface for which the coefficient of rolling friction is μr.
What is the distance that the truck moves up the ramp before coming to a halt? Solve using energy methods.
Express your answer in terms of m , α , v0 , L , g , β and μr .
((v² ÷ 2g) + LSinα) ÷ (Sinβ + µcosβ) is the distance that the truck moves up the ramp before coming to a halt.
Let the distance the truck moves up the ramp be by x.
The kinetic energy of the truck on an icy road is given by,
K1 = (1÷2)mv²
The potential energy of the truck on an icy road is given by,
U1 = mgLSinα
The kinetic energy of the truck on the tuck ramp is given by,
K2 = 0
The potential Energy of the truck-on-truck ramp is given by,
U2 = mgxSinβ
Work done is given by,
W(others) = -µ×mg×cosФ
Hence, by using the work-energy theorem,
W(others) = (K2 + U2)(K1 + U1)
Therefore, by putting the values we get,
((1÷2)mv² + mgLSinα)(0 + mgxSinβ) = -µ×mg×cosФ
x = (K1 + mgLSinα) ÷ (mg(Sinβ + µcosβ))
x = ((v² ÷ 2g) + LSinα) ÷ (Sinβ + µcosβ)
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Find the y-component of this vector.
80.0°
B-18.6 m
0 = 170°
By = [?] m
The y-component of vector B is 3.23 m.
What is y - component of a vector?The y-component of a vector is the vertical component of the vector. It the vector that points in the y-direction.
The y-component of a vector is measured along the y-axis of the Cartesian coordinate.
Mathematically, the y-component of a vector is given as;
By = B sinθ
where;
By is the vertical or y-component of vector BB is the magnitude of vector Bθ is the angle of inclination of vector BThe y-component of vector B is calculated as follows;
By = 18.6 m x sin(170)
By = 3.23 m
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Shiyon just received his first credit card in the mail. His mom reminded him that he needs to order a new pair of baseball cleats. In order to make sure his browser is secure for the transaction, he should
(A) use the same password for all accounts
(B) install Microsoft Excel
(C) enable pop-ups
(D) deactivate background scripts
Answer:
the answer is D
Explanation:
i got it right on the test
In order to make sure his browser is secure for the transaction, Shiyon should deactivate background scripts. Hence, option (D) is correct.
What is background scripts?Without the requirement to create a trigger or script like a business row, a background script provides a free-form method of instantly running server-side code. When you want to clean a modest amount of data that doesn't call for a lot of intricate adjustments, we recommend using a background script.
You could get around that if you need to update records that may be subject to several scripts running on them—as long as they are small or decent in size, not excessive. You'll discover a few techniques to work around that.
Your transaction may run for hours when you run a background script, but this one will end it after four hours.
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3. A parallel-plate capacitor has a capacitance of 1.35 pF. If a 12.0 V battery is connected to this capacitor, how much electrical potential energy would it store?
Answer:
9.72 x 10^(-11)
Explanation:
The energy stored in a capacitor can be calculated as:
[tex]E=\frac{CV^2}{2}[/tex]Where C is the capacitance and V is the Voltage. So, replacing C by 1.35pF or 1.35 x 10^(-12) F, and V by 12.0 V, we get:
[tex]E=\frac{1.35\times10^{-12}F(12.0V)^2}{2}=9.72\times10^{-11}J[/tex]Therefore, the capacitot would store 9.72 x 10^(-11) J of energy.
The electrical potential energy that would be stored in the capacitor is 9.72 x 10^(-11) J. It is the energy which is present in the object when it is at rest.
What is Potential energy?Potential energy is the energy which is stored inside an object which is at rest.
For the parallel plate capacitors, capacitance is dependent upon its geometry, which is given by the formula:
C=ϵ⋅Ad or C = ϵ ⋅ A d ,
where, C is the value of the capacitance,
A is the area of each plate,
d is the distance between the plates, and
ϵ is the permittivity of the material between the plates of the parallel capacitor.
The energy stored in a capacitor can be calculated as:
E = CV²/2
where, C is the capacitance and V is the Voltage.
So, by replacing C with 1.35pF or 1.35 × 10⁻¹²F, and V by 12.0V, we get:
E = 1.35 × 10⁻¹² × (12)²/2
E = 9.72 × 10⁻¹¹ J
Therefore, the capacitor would store 9.72 × 10⁻¹¹ J of energy.
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What in the final position of a robot that starts from (0, 0) m and makes displacements of (4.4) * m (3.3) * m and (- 4, - 2) m?
The final position of the robot is (3, 5) m
Explanation:The initial position of the robot = (0, 0)
The position of the robot after making a displacement of (4, 4)m = (0+4, 0+4)
The position of the robot after making a displacement of (4, 4)m = (4, 4) m
The new position of the robot after making a displacement of (3, 3)m
= (4+3, 4+3)m
The new position of the robot after making a displacement of (3, 3)m =(7, 7)m
The final position of the robot after making a displacement of (-4, -2)m
= (7-4, 7-2)
The final position of the robot after making a displacement of (-4, -2)m
= (3, 5)m
Therefore, the final position of the robot is (3, 5) m
if the mass of the paper is 0.003 kg, what force does the boxer except on it?
Answer:
1.15 N
Explanation:
You want to know the force exerted on a mass of 0.003 kg to accelerate it from 0 to 23 m/s in a period of 0.06 s.
AccelerationThe acceleration of the mass is the change in velocity divided by the change in time:
a = ∆v/∆t
a = ((23 -0) m/s)/(0.06 s) = 1150/3 m/s²
ForceThe force applied is the product of mass and acceleration:
F = ma
F = (0.003 kg)(1150/3 m/s²) = 1.15 kg·m/s² = 1.15 N
The applied force is 1.15 newtons.
Which measurements or observations are needed to calculate density
Answer:
mass and volume measurements are required to calculate density.
A force of 540 N keeps a certain ideal spring stretched a distance of 0.300 m.
A) What is the potential energy of the spring when it is stretched 0.300 m?
B) What is its potential energy when it is compressed 6.00 cm?
A force of 540 N keeps a certain ideal spring stretched a distance of 0.300 m.
A) The potential energy of the spring when it is stretched 0.300 m will be 81 J
B) The potential energy when it is compressed 6.00 cm will be 3.24 J
Hooke's law, law of elasticity discovered by the English scientist Robert Hooke in 1660, which states that, for relatively small deformations of an object, the displacement or size of the deformation is directly proportional to the deforming force or load.
To find spring constant , using formula
k= F/x
where
F = Force
k = spring constant
x = displacement
k = 540 / .300 = 1800 N/m
A) Potential energy of the spring when it is stretched 0.300 m
U = 1/2 * k * [tex]x^{2}[/tex] = 1/2 * 1800 * [tex]0.300^{2}[/tex] = 81 J
B ) Potential energy of the spring when it is stretched 0.06 m
U = 1/2 * k * [tex]x^{2}[/tex] = 1/2 * 1800 * [tex]0.06^{2}[/tex] = 3.24 J
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A rectangular loop of wire with a cross-sectional area of 0.118 m2 carries a current of 0.679 A. The loop is free to rotate about an axis that is perpendicular to a uniform magnetic field strength of 0.179 T. The plane of the loop is initially at an angle of 22.09o to the direction of the magnetic field. What is the magnitude of the torque on the loop ?
Given:
The area of the rectangular loop is,
[tex]A=0.118\text{ m}^2[/tex]The current in the loop is,
[tex]I=0.679\text{ A}[/tex]The magnetic field is of strength
[tex]B=0.179\text{ T}[/tex]The plane of the loop is at an angle
[tex]\theta=22.09\degree[/tex]To find:
The magnitude of the torque on the loop
Explanation:
The torque on a current-carrying loop is,
[tex]\begin{gathered} \tau=IABsin\theta \\ =0.679\times0.118\times0.179\times sin22.09\degree \\ =5.39\times10^{-3}\text{ N.m} \end{gathered}[/tex]Hence, the torque is
[tex]5.39\times10^{-3}\text{ N.m}[/tex]d. Use trigonometry to resolve the following vectors into component
Describe the vectors using vector notation.
v ≈ (3.28512, 20.74146) (3.28512, 20.74146)
In trigonometry, what is a vector?
Vectors can be used to depict motion between points in the coordinate plane. A vector is just a line segment in a specified location with an arrow designating its length and direction. The vectors in the following
Detailed explanationv = 21 (cos (81°), sin (81°))
components for v (3.28512, 20.74146) (x, y)
Authors use different notations. You can write the vector as...
(r, θ) = (21, 81°)
r∠θ = 21∠81°
21 cis 81° = r cis
v = 21·e^(i·9π/20)
(x, y) ≈ (3.28512, 20.74146) (3.28512, 20.74146)
v = 3.28512i +20.74146j Maybe this is the vector notation you're looking for (i and j are unit vectors in the x- and y-directions, respectively)
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Jake throws a rock downward from his deck at time t = 0 and it lands in his yard below.
Which of the following graphs best shows the velocity u of the rock until it hits the ground?
A graph of velocity vs time displays the change in an object's velocity in a straight line with regard to time.
What is meant by Velocity-time graph?A plot between speed and time is called a velocity-time graph. It depicts the motion of an object moving straight ahead. At any given time, the magnitude of velocity equals its instantaneous speed.
A graph of velocity vs time displays the change in an object's velocity in a straight line with regard to time. The x-axis is used to measure time, and the y-axis is used to measure velocity. For an item moving at a constant velocity, the height of the velocity-time graph does not change over time.
We can use a velocity vs. time graph to determine position and a time graph to determine velocity. We are aware that v = d/t. Rearranging the equation with a little mathematics reveals that d = v × t.
The complete question is:
Jack throws a rock downward from his deck at time t = 0 and it lands in his yard below. Make velocity-time graph of the rock until it hits the ground? Assume upward is the positive direction.
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A golf ball is hit horizontally at 40 m/s from the top of a hill that is 2.5 m high. If the terrain around the hill is nearly flat,approximately how far will the golf ball fly? Use - 9.81 m/s2 for the acceleration caused by gravity. Ignore air resistance.Round any intermediate calculations to no less than six decimal places, and round your final answer to two decimal places.
ANSWER
28.56 m
EXPLANATION
Let's make a diagram of this situation to understand it better,
The ball has an initial velocity of 40m/s, which is horizontal. This means that there is no vertical initial velocity.
We have to find the horizontal distance the golf ball flies. Since the horizontal velocity is constant - this is because there is no acceleration in that direction, the distance the ball travels is,
[tex]\Delta x=v_{ox}\cdot t[/tex]The horizontal initial velocity is given, but we have to find the time the ball was in the air. To find it, we use the vertical distance the ball travels - which we know is the height of the hill. In this case, we do have vertical acceleration - the acceleration of gravity, so the vertical distance the ball travels, as shown in the diagram, has a parabolic form and it is given by the equation,
[tex]\Delta y=v_{oy}\cdot t+\frac{1}{2}gt^2[/tex]The initial vertical velocity is zero because the ball is hit horizontally,
[tex]\Delta y=\frac{1}{2}gt^2[/tex]Solve for t. Multiply both sides by 2/g,
[tex]\begin{gathered} \Delta y\frac{2}{g}=\frac{1}{2}g\cdot\frac{2}{g}\cdot t^2 \\ \frac{2\Delta y}{g}=t^2 \end{gathered}[/tex]And take the square root to both sides of the equation,
[tex]t=\sqrt[]{\frac{2\Delta y}{g}}[/tex]Δy is the height of the hill, 2.5m, and g = 9.81m/s²,
[tex]t=\sqrt[]{\frac{2\cdot2.5m}{9.81m/s^2}}=\sqrt[]{\frac{5m}{9.81m/s^2}}=\sqrt[]{0.509684s^2}=0.713922s[/tex]This is the time the ball was in the air for. Now we can find the distance it traveled,
[tex]\Delta x=40m/s\cdot0.713922s=28.56m[/tex]The golf ball flew 28.56 m horizontally.
A 2 kg mass hangs motionless from a spring with spring constant 8 N/m. Howfar will the spring stretch if the mass is motionless?
Set the cannon to have an initial speed of 15 m/s. For which situation do you think the cannon ball will be in the air for the longest time: if it is set at a 60-degree angle, or if it is set at a 70-degree angle?
Question 6 options:
60 degree
70 degree
The cannon ball will be in the air for the longest time if the the angle of projection is set at 70 degrees
How to determine which angle will result in longest time
Case 1:
Initial velocity (u) = 15 m/sAngle of projection (θ) = 60 ° Acceleration due to gravity (g) = 9.8 m/s²Time of flight (T) =?T = 2uSineθ / g
T = (2 × 15 × Sine 60) / 9.8
T = 25.98 / 9.8
T = 2.65 s
Case 2:
Initial velocity (u) = 15 m/sAngle of projection (θ) = 70 ° Acceleration due to gravity (g) = 9.8 m/s²Time of flight (T) =?T = 2uSineθ / g
T = (2 × 15 × Sine 70) / 9.8
T = 28.19 / 9.8
T = 2.88 s
From the above calculations, we can conclude that the ball will stay longer in air if the angle is 70 °
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what does it mean to have a velocity of 10 m/s?
An object of 31.3 kg is at rest. You apply a constant force to this object and the object moves a distance of 103 m in 13 s. find the acceleration produced by your force.
The acceleration produced by your force is determined as 1.22 m/s².
What is the acceleration of the object?
The acceleration of the object is the rate of change of its velocity with time.
Apply the following kinematic equation to solve the acceleration of the object.
s = vt + ¹/₂at²
where;
s is the distance travelled by the objectv is the initial velocity of the object = 0t is the time of motion of the objecta is the acceleration of the objects = ¹/₂at²
103 = ¹/₂a(13)²
103 = 84.5a
a = 103/84.5
a = 1.22 m/s²
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Wei drags a heavy piece of driftwood for 940 m along an irregular path. If Wei ends 710 m from where he starts, exerting a constant force of 625 N parallel to his path the entire time, how much work does he do?
Answer:
587500
Explanation:
Since Wei will still hit every point of the path even though it is irregular, you do not have to worry about the displacement in this problem, just the original path (940m).
W= F x D
587500 = 625N x 940M
3. Find for the total resistance of the series-parallel circuit below
In order to find the total resistance, first let's calculate the total resistance between R2 and R3, which are in series, so the total resistance is the sum:
[tex]R23=R2+R3=3+2=5\text{ ohms}[/tex]Now, let's calculate the total resistance between R23 and R1, which are in parallel, so we use the equation below:
[tex]R123=\frac{R23\cdot R1}{R23+R1}=\frac{5\cdot6}{5+6}=\frac{30}{11}=2.727\text{ ohms}[/tex]Finally, we calculate the total resistance between R123 and R4, which are in series:
[tex]RT=R4+R123=3+2.727=5.727\text{ ohms}[/tex]thunder can make objects inside a room vibrate explain what causes the objects to vibrate
Thunder can make objects inside a room vibrate by which our house columns in underground that causes the objects to vibrate.
What causes the sound of thunder?Thunder is caused by the rapid expansion of air around the path of lightning. It only takes a few millionths of a second for lightning to burst through the air from a cloud to a nearby tree or roof. The loud thunder that follows a lightning strike is generally said to come from the lightning itself. However, the rumbling and roaring heard during thunderstorms is actually due to the rapid expansion of air around the lightning.
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An Earth satellite moves in circular orbit 602 km above the earths surface with a period of 96.53 min. What are the speed and the magnitude of the centripetal acceleration of the satellite.
The speed and the magnitude of the centripetal acceleration of the satellite are 7.568 km/s and 8.205 x 10^-3 m / s^2
Here this problem we are dealing with the speed and centripetal acceleration where the speed of the satellite is the speed required to attain adjustment between gravity's pull on the satellite and the inertia of the satellite's motion whereas centripetal acceleration is referred to as the acceleration of a body navigating a circular way.Since we are given the orbital radius which is 602 km and the period which is 96.53 min.
The radius of the satellite's orbit, r = earth radius + orbit radius
=>r =6378 km + 602 km = 6980 km
Since the formula for the orbital circumference is :
2πR
=>2 • 3.14 •6980 = 43,834.4 km
Now divide the circumference by the time period we get the orbital speed which is,
43,834.4 km / (96.53 min • 60) = 7.568 km/s
the formula we are referring for calculating the centripetal acceleration is:
a = v²/ r ,
=> a = (7.568 ^2) / 6980 = 8.205 x 10^-3 m / s^2
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A boat is moving across a 100 meter wide stream at 5 m/s. If the stream is flowing south at 10 m/s, how long will it take the boat to cross the river?
answer:
20 minutes for 5 m/s 10 minutes for 10 m/s
Which of the following statements is always true?
a.
Pressure always increases when force increases or the area acted on increases.
b.
Pressure always increases when force increases or the area acted on decreases.
c.
Pressure always increases when force decreases or the area acted on increases.
d.
Pressure always increases when force decreases or the area acted on decreases
The statement that is true is that pressure always increases when force increases or the area acted on decreases (option B).
What is pressure?Pressure is the amount of force that is applied over a given area divided by the size of this area.
The pressure applied to a surface can be calculated by dividing the force required by the area of the surface as follows:
P = F/A
The above shows that the pressure of a body is inversely proportional to the area but directly proportional to the force.
This means that as the pressure of an object increases, the force increases while the area acted on decreases.
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A person walks 8 km east, turns and walks 5 km west, then turns back around and walks 3 km, east.
What is the distance traveled of the person from the initial point to ending point?
A person walks 8 km east, turns and walks 5 km west, then turns back around and walks 3 km, east. Total distance traveled of the person from the initial point to ending point will be 16 km
Distance is the total movement of an object without any regard to direction. We can define distance as to how much ground an object has covered despite its starting or ending point.
total distance = 8 km + 5 km + 3 km = 16 km
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Please help me
how law of motion applied in basketball explain
Explanation:
According to the first law of motion, the basketball is always moving in one direction, unless acted on by another force.
Jason is pulling a box across the roompulling with a force of 16 newtons and his arm is making a 68 anglet with the horizontal what is the horizontal component of the he is pulling with?
The horizontal component of a force is defined as follows
[tex]F_x=F\cdot\cos \theta[/tex]Where F = 16 N, and theta = 68. Let's use these values to find the horizontal component of F.
[tex]F_x=16N\cdot\cos 68\approx6N[/tex]Therefore, the horizontal component of the pulling force is 6 Newtons.Provide an example of a situation that can be explained by each of Newton's Three Laws of Motion.
ANSWER and EXPLANATION
First, let us state Newton's three laws of motion.
The first law of motion states that:
The second law of motion states that:
The third law of motion states that:
Let us take an example of a ball that is stationary on a frictionless floor. As long as no force acts on the ball, it will continue to remain at rest but once a person kicks the ball, it begins to move and stay in motion (since the floor is frictionless, hence, no resistive force).
That is an application of the first law of motion.
The acceleration of this ball depends on the force with which it was kicked and the mass of the ball, as shown in the force equation:
[tex]F=ma[/tex]where m = mass, a = acceleration
That is an application of the second law of motion.
Assuming that this ball then comes in contact with a wall and stops moving. The force that the ball exerts on the wall is equal to the force that the wall exerts on the ball, but both forces act opposite one another.
That is an application of the third law of motion.
A 50-ω resistor is connected to a 9.0 V battery. How much thermal energy is produced in 7.5 minutes?1.2 102 J1.3 103J3.0 102 J7.3 102 J
The power consumed by the resistor can be expressed as,
[tex]P=\frac{V^2}{R}[/tex]Substitute the given values,
[tex]\begin{gathered} P=\frac{(9.0V)^2}{50\text{ }\Omega}(\frac{1\text{ W}}{1V^2\Omega^{-1}}) \\ =1.62\text{ W} \end{gathered}[/tex]The thermal energy produced in the given time can be expressed as,
[tex]E=Pt[/tex]Plug in the known values,
[tex]\begin{gathered} E=(1.62\text{ W)(7.5 min)(}\frac{60\text{ s}}{1\text{ min}}) \\ =729\text{ J} \\ \approx7.3\times10^2\text{ J} \end{gathered}[/tex]Thus, the thermal energy produced is 7.3*10^2 J which means fourth option is correct.