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2 years ago

11

A 64.5 kg person steps off a 129 kg rowboat with a force of 34.0 N. What is the force that is applied to the person by the rowbo

at?

1 answer:

Arada [10]2 years ago

3 0

34.0 N which is the force that applied to the person by rowboat.

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A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp

liq [111]

Answer:

$W_2=\sqrt{\frac{3}{5} }W_1$

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

$I_1 =\frac{2}{5}MR^2$

$K_1 = \frac{1}{2}IW_1^2$

So, replacing, we get that:

$K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

At the same way, the moment of inertia and kinetic energy for second ball is:

$I_2 =\frac{2}{3}MR^2$

$K_2 = \frac{1}{2}IW_2^2$

So:

$K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2$

Then, $K_2$ is equal to $K_1$ , so:

$K_2 = K_1$

$\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

$\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2$

$\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2$

Finally, solving for $W_2$ , we get:

$W_2=\sqrt{\frac{3}{5} }W_1$

5 0

2 years ago

If a rock is thrown upward on the planet mars with a velocity of 14 m/s, its height (in meters) after t seconds is given by h =

crimeas [40]

<u>Answer:</u>

Velocity of rock after 2 seconds = 6.56 m/s

<u>Explanation:</u>

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Here height of rock in meters, h = $14t-1.86t^2$

Comparing both the equations

We will get initial velocity = 14 m/s(already given) and $\frac{1}{2} a = -1.86$

So, Acceleration, a = -3.72 $m/s^2$

Now we have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

When time is 2 seconds we need to find final velocity.

v = 14 - 3.72 * 2 = 6.56 m/s.

So, Velocity of rock after 2 seconds = 6.56 m/s

6 0

2 years ago

The gravitational force produce between any two object kept 2.5×10 to the power 4 km apart is 580N.At what distance should they

timofeeve [1]

Answer:

d = 3.54 x 10⁴ Km

Explanation:

Given,

The distance between the two objects, r = 2.5 x 10⁴ Km

The gravitational force between them, F = 580 N

The gravitational force between the two objects is given by the formula

F = GMm/r² newton

When the gravitational force becomes half, then the distance between them becomes

Let us multiply the above equation by 1/2 on both sides

( 1/2) F = (1/2) GMm/r²

= GMm/2r²

= GMm/(√2r)²

Therefore, the distance becomes √2d, when the gravitational force between them becomes half

d = √2r = √2 x 2.5 x 10⁴ Km

= 3.54 x 10⁴ Km

Hence, the two objects should be kept at a distance, d = 3.54 x 10⁴ Km so that the gravitational force becomes half.

3 0

2 years ago

Which of the following statements are true about an object in two-dimensional projectile motion with no air resistance? (There c

ki77a [65]

Answer:

The correct answers are

The following statements are true about an object in two-dimensional projectile motion with no air resistance

D) The speed of the object is zero at its highest point.

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

Explanation:

A) The speed of the object is constant but its velocity is not constant.

False the vertical velocity increases on descent

B) The acceleration of the object is constant but its object is + g when the object is rising and -g when it is falling.

False, the acceleration is -g when the object is rising

C) The acceleration of the object is zero at its highest point.

False, the acceleration is constant in magnitude throughout the motion

D) The speed of the object is zero at its highest point.

True, the direction of motion changes at the highest point from hence the body comes to rest and the speed is zero

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

True, the horizontal acceleration has associated force during motion but the vertical acceleration is due to gravity which is constant downwards

6 0

2 years ago

30) A force produces power P by doing work W in a time T. What power will be produced by a force that does six times as much wor

schepotkina [342]

Answer:

A) 12P

Explanation:

The power produced by a force is given by the equation

$P=\frac{W}{T}$

where

W is the work done by the force

T is the time in which the work is done

At the beginning in this problem, we have:

W = work done by the force

T = time taken

So the power produced is

$P=\frac{W}{T}$

Later, the force does six times more work, so the work done now is

$W'=6W$

And this work is done in half the time, so the new time is

$T'=\frac{T}{2}$

Substituting into the equation of the power, we find the new power produced:

$P'=\frac{W'}{T'}=\frac{6W}{T/2}=12\frac{W}{T}=12P$

So, 12 times more power.

4 0

2 years ago