Answer:
a) 4.0 rad/s2
Explanation:
For rigid bodies, Newton's 2nd law becomes :τ = I * α (1)
where τ is the net external torque applied, I is the rotational inertia
of the body with respect to the axis of rotation, and α is the angular
acceleration caused by the torque.
At the same time, we can apply the definition of torque to the left side of (1), as follows:[tex]\tau = F*r*sin \theta (2)[/tex]
where τ = external net torque applied by Fnet, r is the distance
between the axis of rotation and the line of Fnet, and θ is the
angle between both vectors.
In this particular case, as Fnet is applied tangentially to the disk, Fnet
and r are perpendicular each other.
Since left sides of (1) and (2) are equal each other, right sides are equal too, so we can solve for the angular acceleration as follows:[tex]\alpha = \frac{F*r}{I} = \frac{5.0N*1.6m}{2.0 kg*m2} = 4.0 rad/s2 (3)[/tex]
8) a 20kg box is sliding across the ground. If the coefficient of friction is 0.5, how much friction will the box experience?
A 490 N
B 19.6 N
C 98 N
D 2 N
Answer:
STOP CHETING
Answer:
C is the answer to your question
Explanation:
Have a great day!!! :)
A 80 kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 5 meters starting from rest, its speed is 6.0 m/s. Find the magnitude of the net force on the bobsled.
How do you solve this question?
Answer:
F = 288 [N]
Explanation:
To solve this problem we must use the following equation of kinematics and find the value of acceleration.
[tex]v_{f}^{2} =v_{o}^{2} +2*a*x[/tex]
where:
Vf = final velocity = 6 [m/s]
Vo = initial velocity = 0 (starting from rest)
a = acceleration [m/s²]
x = distance = 5 [m]
Now replacing, we have:
[tex](6)^{2}=0+(2*a*5)\\36=10*a\\a = 3.6 [m/s^{2}][/tex]
Since we already have the value of acceleration, we can use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
ΣF = m*a
[tex]F =80*3.6\\F = 288 [N][/tex]
Does anyone knows this for physics?
A rocket falls from the apogee (0 meters per second) until it hits the ground with a speed of 10 meters per second. Gravity pulled it down with an acceleration of 9.8m/s^2. The time during which the ball is in free fall is approximately what time?
Answer:
Approximately 1.02 seconds
Explanation:
Use the final velocity (vf) formula for a uniformly accelerated movement under "g" (acceleration of gravity):
[tex]v_f=v_i+g\,*\,t[/tex]
in our case:
[tex]10=0+9.8\,*\,t\\t=10/9.8\\t\approx 1.02\,\,sec[/tex]
initial velocity 10 m/s accelerates at 5 m/s for 2 seconds whats the final velocity
Answer:
The final velocity is 20 m/s.
Explanation:
Constant Acceleration Motion
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:
[tex]v_f=v_o+at[/tex]
The provided data is: vo=10 m/s, [tex]a=5\ m/s^2[/tex], t=2 s. The final velocity is:
[tex]v_f=10~m/s+5\ m/s^2\cdot 2\ s[/tex]
[tex]v_f=20\ m/s[/tex]
The final velocity is 20 m/s.
Steelhead trout migrate upriver to spawn. Occasionally they need to leap up small waterfalls to continue their journey. Fortunately, steelhead are remarkable jumpers, capable of leaving the water at a speed of 8.0 m/s. What is the maximum height that a steelhead can jump
Answer:
s = 3.26 m
Explanation:
Given that,
Water leaves at a speed of 8 m/s
We need to find the maximum height that steelhead can jump. Let it can jump to a height of h.
At maximum height, final speed is equal to 0. We can use third equation of motion to find the maximum height.
[tex]v^2-u^2=2as[/tex]
a = -g
[tex]-u^2=-2gs\\\\s=\dfrac{u^2}{2g}\\\\s=\dfrac{(8)^2}{2\times 9.8}\\\\=3.26\ m[/tex]
Hence, the maximum height is 3.26 m.
If Jack weighs more that Jill, and they run up the same hill, who has done more work?
If Jack and Jill weigh the same, and Jill runs up the hill in half the time as Jack, who had more power?
Answer:
jack has done more work pulling more weight and Jill has more power.
Explanation:
A projectile is fired horizontally from a height of 10 m above level ground. The projectile lands a horizontal distance of 15 m from where it was launched.
-Find the hang time for the projectile.
-Find the initial speed of a projectile.
-What are the x and y components of the projectile’s velocity the moment before it strikes the ground?
-At what speed will the projectile strike the ground?
Answer:
a)t = 1,43 s
b) V = 10,49 m/s
c) V₀ₓ = 10,49 m/s ; V₀y = 14,01 m/s
d) Vf = 17,5 m/s
Explanation:
According to the problem statement
V₀ = V₀ₓ and V₀y = 0
And at the end of the movement t = ? the distance y = 10 m
Therefore as
h = V₀y - (1/2)*g*t²
Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²
10 = 4,9*t²
t² = 10/4,9 ⇒ t² = 2,04 s
t = 1,43 s
a) 1,43 s is the time of movement
b) V₀ = V₀ₓ V₀y = 0 and V₀ₓ = Vₓ ( constant )
Just before touching the ground, the horizontal distance is
hd = 15 = Vₓ * t
Then 15 /1,43 = Vₓ = V₀ₓ
Vₓ = 10,49 m/s
Then initial speed is V = 10,49 m/s since V₀y = 0
Vf² = Vₓ² + Vy²
Vyf = V₀y - g*t
Vyf = 0 - 9,8 *1,43
Vyf = - 14,01 m/s
And finally the speed when the projectile strike the ground is:
Vf² = Vₓ² + Vy²
Vf = √ (10,49)² + (14,01)²
Vf = 17,50 m/s
What is the formula for electrical power?
A. P = IV/R
B. P = IVR
C. P = IR
D. P = IV
Correct answer is D!
Electrical power can be defined as the product of Voltage and current. Thus, P = IV. Hence, Option D is the correct answer.
What is Power?As we study Electric power, the electric current to the another form of energy.The SI unit of power is the watt or one joule per second.Electric power can be calculated as current times voltage. Electrical energy used by equals the power of the appliances and its multiplied by the amount of time and the appliance is used.Thus, the formula for electric power,
P = IV watt
Where,
P - Power (Watt)
I - Current (Ampere)
V - Voltage (volts)
Power can be also known as the mechanical power equation.Also, the power dissipated energy can also be found with the equation.Electrical power can be defined as the product of Voltage and current. Thus, P = IV. Hence, Option D is the correct answer.
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A rescue helicopter is lifting a man with 86 kg mass from a capsized boat by means of a cable and harness. What is the tension in the cable when the man is given an initial upward acceleration of 1.35 m/s2
The man is pulled upward by a net force of
∑ F = (86 kg) (1.35 m/s²) = 116.1 N
The net force is comprised of the tension T in the cable and the man's weight W, such that
∑ F = T - W = 116.1 N
T = (86 kg) (9.80 m/s²) + 116.1 N
T = 958.9 N ≈ 960 N
A drag racing vehicle travels from 0 to 100 mph in 5 seconds north. What is the acceleration?
Answer:
20 m/s^2
Explanation:
Use the formula Vf=Vi+at
Vf=100
Vi=0
a=?
t=5
Then solve for (a)
what instrument is used to measure volume by displacement
The fact that our preconceived ideas contribute to our ability to process new information best illustrates the importance of: the serial position effect. O repression iconic memory . semantic encoding . retroactive interference .
Answer:
It’s a
Explanation:
Don’t actually put that i needed the points mb
Which term describes the force applied to an object multiplied by the time the force is applied
Answer:
Momentum
Explanation:
In order for an object's momentum to change,a force must be applied for a period of time
which of the following is a lever
Answer: the last one
Explanation: it pushes the lid off in a upward motion
Answer:
the last one
Explanation:
Can you mark me as BRAINLLIEST, PLEASE!!!!!!!!!!!!
(○’ω’○) (ToT) ↖(^ω^)↗ -_-z ◑﹏◐ 囧 ^O^ B-)
The diagram shows an electromagnet made with copper wire, a steel nail,
and a 1.5 V battery. Which action could cause this electromagnet to be
stronger?
A. Replace the steel nail with a plastic straw.
B. Replace the battery with a 6 V battery.
C. Reduce the number of coils of wire wrapped around the nail.
D. Reverse the direction of the battery in the circuit.
Correct answer is B!
Answer:B
Explanation: i took the test and i got it right
If 710-nm and 655-nm light passes through two slits 0.65 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.1 m away?
Answer:
Explanation:
Position of n the order fringe = n λ D / d
for n = 2
position = 2 λ D / d
λ = 710 nm , D = 1.1 m
d = .65 x 10⁻³
position 1 = 2 x 710 x 10⁻⁹ x 1.1 / .65 x 10⁻³
= 2403.07 x 10⁻⁶ m
= 2.403 x 10⁻³ m
= 2.403 mm .
For λ = 655 nm
position = 2 λ D / d
λ = 655 nm , D = 1.1 m
d = .65 x 10⁻³
position 2 = 2 x 655 x 10⁻⁹ x 1.1 / .65 x 10⁻³
= 2216.91 x 10⁻⁶ m
= 2.217 x 10⁻³ m
= 2.217 mm .
Difference between their position
= 2.403 - 2.217 = .186 mm .
1. The property of a material to absorb liquid like water is called
A non-porous
C. buoyancy
B. density
D. porosity
HELP ME PLEASE I MARK BRAINLIESTIF YOU ANSWER THIS♡
Answer:
A hygroscopic substance is one that readily attracts water from its surroundings, through either absorption or adsorption.
Explanation:
A hygroscopic substance is one that readily attracts water from its surroundings, through either absorption or adsorption.
A particle executes simple harmonic motion with an amplitude of 2.00 cm. At what positions does its speed equal one fourth of its maximum speed?
Answer:
The positions are 0.0194 m and - 0.0194 m.
Explanation:
Given;
amplitude of the simple harmonic motion, A = 2.0 cm = 0.02 m
speed of simple harmonic motion is given as;
[tex]v = \omega \sqrt{A^2-x^2}[/tex]
the maximum speed of the simple harmonic motion is given as;
[tex]v_{max} = \omega A[/tex]
when the speed equal one fourth of its maximum speed
[tex]v =\frac{v_{max}}{4}[/tex]
[tex]\omega\sqrt{A^2-x^2} = \frac{\omega A}{4} \\\\\sqrt{A^2-x^2}= \frac{A}{4}\\\\A^2-x^2 = \frac{A^2}{16} \\\\x^2 = A^2 - \frac{A^2}{16} \\\\x^2 = \frac{16A^2 - A^2}{16} \\\\x^2 = \frac{15A^2}{16} \\\\x= \sqrt{\frac{15A^2}{16} } \\\\x = \sqrt{\frac{15(0.02)^2}{16} }\\\\x = 0.0194 \ m \ \ or\ - 0.0194 \ m[/tex]
Thus, the positions are 0.0194 m and - 0.0194 m.
The positions where the speed equals 1/4 of its maximum speed is mathematically given as
[tex]x \pm 0.0194[/tex]
What positions does its speed equal one-fourth of its maximum speed?Question Parameters:
an amplitude of 2.00 cm.
Generally, the equation for the speed of simple harmonic motion is mathematically given as
[tex]v = w \sqrt{A^2-x^2}\\\\v=\frac{wA}{4}[/tex]
Therefore
[tex]w \sqrt{A^2-x^2}=\frac{wA}{4}\\\\x^2=A^2-A^2/16\\\\x=\sqrt{\frac{15A^2}{16}}[/tex]
[tex]x \pm 0.0194[/tex]
In conclusion, the positions are
[tex]x \pm 0.0194[/tex]
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in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s
Answer:25 N
Explanation:
in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s is 25N
what is speed ?Speed is the ratio of distance with respect to the time in which the distance was covered. Speed is a scalar quantity as it does not have magnitude only have direction
The formula of speed can be represented as s=d/t, Where, s is the speed in m.s-1, d is the distance traveled in m, t is the time taken in s
Uniform speed is defined when the object covers equal distance at equal time intervals, variable speed is defined as when the object covers a different distance at equal intervals of times.
Average speed is defined as the total distance travelled by an object to the total time taken by the object.
Instantaneous speed is defined as when the object is move with variable speed, then the speed at any instant of time is known as instantaneous speed.
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Answer this question: Is math really important to giving science power? (remember the 5 Ws and the H)
A horse has a momentum of 1200 kg·m/s. If the horse has a mass of 313 kg, what is the speed of the horse?
Answer:
3.83 m/sExplanation:
The speed of the horse can be found by using the formula
[tex]v = \frac{p}{m} \\ [/tex]
p is the momentum
m is the mass
From the question we have
[tex]v = \frac{1200}{313} \\ = 3.83386..[/tex]
We have the final answer as
3.83 m/sHope this helps you
a 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Find the total momentum of the system.
Explanation:It is given that,
Mass of bumper car, m₁ = 202 kg
Initial speed of the bumper car, u₁ = 8.5 m/s
Mass of the other car, m₂ = 355 kg
Initial velocity of the other car is 0 as it at rest, u₂ = 0
Final velocity of the other car after collision, v₂ = 5.8 m/s
Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁
Using the conservation of linear momentum as :
p₁ = m₁v₁ = -342 kg-m/s
So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.
Answer:
1717 kg*m/s
Explanation:
found it put it in on acellus and it was right
The students look through the side of the aquarium.
They notice that the image of the tongs appears to break as the tongs enter the water.
Which property of light are the students observing in this situation?
Answer:
light refraction
Explanation:
A ball is thrown off a cliff at a speed of 10 m/s in a horizontally direction. The ball reaches the ground 1.5 seconds. If the ball is launched a second time at the same speed from a second higher cliff, which of the following is true?
A. The ball takes a longer time and lands further away from the foot of the cliff.
B. The ball takes longer to hit the ground, but lands at the same distance from the foot of the cliff.
C. The ball takes the same time lands at the same distance from the foot of the cliff.
D. The ball falls further away from the foot of the cliff, but takes the same time.
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Why did the bowling ball make a bigger
splash than the ping pong ball?
What kind of energy made that splash
happen?
Answer:
a. The bowling ball would have more kinetic energy because of its greater mass.
b. Potential energy
Explanation
a. Bowling ball has higher mass, self explanatory.
b. A high diver has lots of stored energy when they are on the diving platform. When they dive this stored energy helps make the splash when they hit the water. Stored energy is also called potential energy.
While getting buff at the gym you lift a bunch of weights applying 1000N of force to lift them from the ground to a height of 2m. How much work did you do?
A. 2000 J
B. 1000J
C. -2000 J
D. -1000 J
Which of the following is true?
A
The Atlantic, Pacific, Indian, Arctic, and Southern Oceans are completely separate
from each other.
B
The ocean covers about half of the Earth's surface.
с
Scientists have studied most of the ocean, but a tiny bit remains unexplored.
D
Scientists know more about the moon than they do the ocean.
Answer:
options B,C,D are true
Explanation:
What happens to the sunlight that does not reach Earth’s surface?
A block with mass m = 0.200 kg is placed against a compressed spring at the bottom of a ramp that is at an angle of 53.0∘ above the horizontal. The spring has 8.00 J of elastic potential energy stored in it. The spring is released, and the block moves up the incline. After the block has traveled a distance of 3.00 m, its speed is 4.00 m/s. Part A What is the magnitude of the friction force that the ramp exerts on the block while the block is moving? Express your answer with the appropriate units.
Answer:
[tex]0.566\; \rm N[/tex] (assuming that while the block is moving, the friction on the block is constant.)
Explanation:
The mechanical energy of a system is the sum of its:
elastic potential energy, gravitational potential energy, andkinetic energy.Friction does work on the block as the block moves up the ramp. The amount of energy that the block-string system has lost would be equal in size to the work that friction has done on the block. The size of the friction on the block could thus be computed.
Before this block was released, the block-spring system has no kinetic energy because there was no movement. Assume that the system has no gravitational potential energy at that moment, either. The only type of mechanical energy in this system at that moment would be elastic potential energy: [tex]8\; \rm J[/tex] according to the question.
The question states that the ramp is [tex]53.0^\circ[/tex] above horizontal. Therefore, after the block has traveled [tex]3.00\; \rm m[/tex] (along the ramp,) the height of this block would have increased by [tex]\Delta h = 3.00\; \rm \sin\left(53^\circ\right) \approx 2.39591\; \rm m[/tex]. Calculate the corresponding gain in gravitational potential energy:
[tex]\begin{aligned}& m \cdot g \cdot \Delta h\\ &\approx 0.200\; \rm kg& \\\ &\quad\times 9.81\; \rm N \cdot kg^{-1} \\ &\quad \times 2.39591\; \rm m \\ &\approx 4.70077\; \rm J\end{aligned}[/tex].
On the other hand, the question states that the speed of the block ([tex]m = 0.200\; \rm kg[/tex]) at that moment is [tex]v = 4.00\; \rm m \cdot s^{-1}[/tex]. Calculate the corresponding kinetic energy:
[tex]\begin{aligned}& \frac{1}{2}\, m \cdot v^{2} = 1.6\; \rm J \end{aligned}[/tex].
Because the block was already released, there should be no elastic potential energy in the spring.
Hence, the mechanical of the block-spring system would be approximately [tex]4.70077\; \rm J+ 1.6\; \rm J \approx 6.30077\; \rm J[/tex].
Approximate the amount of mechanical energy that is lost:
[tex]8\; \rm J - 6.30077\; \rm J \approx 1.69923\; \rm J[/tex].
In other words, when applied over [tex]3\; \rm m[/tex], the friction on this block would do approximately [tex]1.69923\; \rm J[/tex] of work. Approximate the size of that friction:
[tex]\begin{aligned}F &= \frac{W}{s} \\ &\approx \frac{1.69923\; \rm J}{3.00\; \rm m}\approx 0.566\; \rm N\end{aligned}[/tex].