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4 years ago

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Steven carefully places a ????=1.95 kgm wooden block on a frictionless ramp so that the block begins to slide down the ramp from

rest. The ramp makes an angle of ????=54.3° up from the horizontal. Which forces do nonzero work on the block as it slides down the ramp?

1 answer:

NemiM [27]4 years ago

4 0

Answer:

The forces that do non-zero work on the block are gravity and normal reaction force

Explanation:

In this question we have given,

mass of the wooden block, $m=1.95 kg$

angle between ramp and horizontal surface, $\alpha =54.3^o$

in this case ramp is friction less so friction force will be zero and hence work done due to friction force will be zero.

As wooden block is sliding down due to gravity(gravity is pulling the block in downward direction).

so two forces will act on wooden block as it slides down the ramp that are gravity and normal reaction force.

Therefore, the forces that do non-zero work on the block are gravity and normal reaction force.

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Two 30 uC charges lie on the x-axis, one at the origin and the other at 9 m. A third point is located at 27 m. What is the poten

alukav5142 [94]

Answer:

25000 V

Explanation:

The formula for potential is

V = Kq/r

Potential at B due to the charge placed at origin O

V1 = K q / OB

$V_{1}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{27}$

V1 = 10000 V

Potential at B due to the charge placed at A

V2 = K q / AB

$V_{2}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{18}$

V2 = 15000 V

Total potential at B

V = V1 + V2 = 10000 + 15000 = 25000 V

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3 years ago

A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2. How long does it take to come to rest? How many

Darya [45]

Answer:

$t=45.7s$

$\alpha =116revolutions$

Explanation:

Since we have given values of ω₀=32.o rad/s ,ω=0 and α=-0.700 rad/s² to find t we use below equation

$w=w_{o}+at\\ 0=(32.0rad/s)+(-0.700rad/s^{2} )t\\t=\frac{-32.0}{-0.700} \\t=45.7s$

To find revolutions we use below equation

$w^{2}=w_{o}^{2}+2a\alpha$

Substitute the given values to find revolutions α

So

$0=(32.0rad/s)^{2}+2(-0.700rad/s^{2} )\alpha \\\alpha =\frac{(-32.0rad/s)^{2}}{2(-0.700rad/s^{2} )} \\\alpha =731rad$

To convert rad to rev:

$\alpha =(731rad)*(\frac{1rev}{2\pi rad} )\\\alpha =116revolutions$

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4 years ago

In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give

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Answer:

The current pass the $R_2$ is $I = 0.25 A$

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The voltage is $V = 3V$

The first resistance is $R_1 = 7 \Omega$

The second resistance is $R_2 = 5 \Omega$

Since the resistors are connected in series their equivalent resistance is

$R_{eq} = R_1 +R_2$

Substituting values

$R_{eq} = 7 + 5$

$R_{eq} = 12 \Omega$

Since the resistance are connected in serie the current passing through the circuit is the same current passing through $R_2$ which is mathematically evaluated as

$I = \frac{V}{R_{eq}}$

Substituting values

$I = \frac{3}{12}$

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4 years ago

When a 9.9 kg object is in free fall, it feels a force of 97.02 N. what is its acceleration?

Free_Kalibri [48]

The acceleration of an object in free fall is $9.8 m/s^2$

Explanation:

We can solve the problem by using Newton's second law of motion, which states that:

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F is the net force on an object

m is the mass of the object

a is its acceleration

For the object in this problem, we have:

m = 9.9 kg is its mass

F = 97.02 N is the net force on the object

So, we can rearrange the equation to find its acceleration:

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This value is actually known as acceleration of gravity (g), and it is the acceleration that any object in free fall has near the Earth's surface, regardless of its mass.

Learn more about Newton's second law here:

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When the grasshoppers vertical velocity is exactly zero.
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g: 9.81
t: time you are looking for
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3 years ago

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