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3 years ago

15

Starting from rest, a 5.00 kg steel sphere rolls down a frictionless ramp with a height of 4.00 m. What is the sphere's speed wh

en it reaches the bottom of the ramp? (Use 9.80 for the value of "g.")

1 answer:

elena-14-01-66 [18.8K]3 years ago

8 0

Answer:

8.85437 m/s

Explanation:

m = Mass of sphere = 5 kg

h = Vertical height = 4 m

g = Acceleration due to gravity = 9.80 m/s²

Applying conservation of energy we get

$\dfrac{1}{2}mv^2=mgh$

$\\\Rightarrow v=\sqrt{2gh}$

$\\\Rightarrow v=\sqrt{2\times 9.8\times 4}$

$\\\Rightarrow v=8.85437\ m/s$

The sphere's speed when it reaches the bottom of the ramp is 8.85437 m/s

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Find the instantaneous acceleration at t=ls for an object moving along a straight axis with velocity function:

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The answer is A.
Explanation:
We know that the average acceleration a for an interval of time Δt is expressed as:

a = Δv
Δt
where Δv is the change in velocity that occurs during Δt.
e formula for the instantaneous acceleration a is almost the same, except that we need to indicate that we're interested in knowing what the ratio of Δv to Δt approaches as Δt approaches zero.

We can indicate that by using the limit notation.

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3 years ago

A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an

anyanavicka [17]

Answer:

The value of tangential acceleration $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

The value of radial acceleration $\alpha_{r} = 80 \frac{m}{s^{2} }$

Explanation:

Angular acceleration = 50 $\frac{rad}{s^{2} }$

Radius of the disk = 0.8 m

Angular velocity = 10 $\frac{rad}{s}$

We know that tangential acceleration is given by the formula $\alpha_{t} =$ $r \alpha$

Where r = radius of the disk

$\alpha$ = angular acceleration

⇒ $\alpha_{t} =$ 0.8 × 50

⇒ $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

This is the value of tangential acceleration.

Radial acceleration is given by

$\alpha_{r} = \frac{V^{2} }{r}$

Where V = velocity of the disk = r $\omega$

⇒ V = 0.8 × 10

⇒ V = 8 $\frac{m}{s}$

Radial acceleration

$\alpha_{r} = \frac{8^{2} }{0.8}$

$\alpha_{r} = 80 \frac{m}{s^{2} }$

This is the value of radial acceleration.

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4 years ago

A car moving at 11.6 m/s begins to accelerate at 5.22 m/s/s and reaches a final velocity of 31.5 m/s. How far did it travel duri

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Use the kinematic equation,

Vf^2 = Vi^2 + 2aX

31.5^2 = 11.6^2 + 2(5.22)(x)

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3 years ago

A 6 kg block is sliding down a horizontal frictionless surface with a constant speed of 5 m/s. It then slides down a frictionles

BARSIC [14]

Answer:

5in

Explanation:

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4 years ago

In a circus act, a 70 kg clown is shot from a cannon with an initial velocity of 17 m/s at some unknown angle above the horizont

EleoNora [17]

Answer:

$K_{2}=7302.4J$

Explanation:

Given the initial velocity of the clown, his mass and final height we can calculate the final kinetic energy using the <em><u>conservation of total mechanical energy</u></em>

$K_{1}+U_{1}=K_{2}+U_{2}$

$K_{2}=K_{1}+U_{1}-U_{2}$

$K_{2}=\frac{1}{2}mv_{1}^{2}+mgh_{1}-mgh_{2}$

Since $h_{1}=0$

$K_{2}=\frac{1}{2}mv_{1}^{2}-mgh_{2}$

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4 years ago