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statuscvo [17]
2 years ago
11

A 2.0kg solid disk rolls without slipping on a horizontal surface so that its center proceeds to the right with a speed of 5.0 m

/s. What is the instantaneous speed of the point of the disk that makes contact with the surface?
Physics
1 answer:
Alexandra [31]2 years ago
4 0

Answer:

Instantaneous speed of contact point will be ZERO

Explanation:

As we know that disc is rolling without slipping on horizontal surface

So here the speed of center of the disc is given as

v = 5 m/s

now at the contact point the tangential speed will be in reverse direction

v_t = R\omega

now we know that net contact speed with respect to its lower surface must be zero

v_{net} = v - v_t = 0

so net velocity of contact point with respect to its lower surface must be ZERO here

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A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w
Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

4 0
1 year ago
Mr. Smith is designing a race where velocity will be measured. Which course would allow velocity to accurately get a winner?
liraira [26]
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6 0
2 years ago
Read 2 more answers
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
2 years ago
abin is doing work by lifting a bowling ball. Which statement could be made about the energy in this situation?
PtichkaEL [24]
The statement that could be made about the energy in this situation would be :
It being transferred from his arms muscles to the ball.

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8 0
2 years ago
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Some vehicles, like _____, have different rules for registration and renewal. A. trailers B. coupes C. sedans D. pickup trucks
ycow [4]

Answer:

trailers

Explanation:

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