Suppose you exert a 25-N force to lift a ball 0.4 m in 2 s. How much work is done?

Formulating a Hypothesis: Part I Since the investigative question has two variables, you need to focus on each one separately. T

Westkost [7]

Answer : The kinetic energy depends directly on the mass of a particle.

Explanation :

We know that the kinetic energy of any particle is given by :

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

Where,

m is the mass of an object.

v is the velocity with which it is moving

Kinetic energy is due to the motion of the particle.

So, the kinetic energy of a particle is directly proportional to its mass.

Hence, the conclusion of the question is if the mass of a particle is increases then its kinetic energy also increase.

3 0

3 years ago

Read 2 more answers

Tarzan and Jane. Because of your concern that incorrect science is being taught to children when they watch cartoons on TV, you

creativ13 [48]

Answer:

The maximum height Tarzan and Jane can swing as a fraction of her initial heigh is ⅓h

Explanation:

Let

m = Mass of Tarzan

M = Mass of Jane

Given

M = 2m

To calculate the maximum height Tarzan and Jane can swing, we make use of the potential energy at their initial and final position.

Reason being that;

At both the initial and final position, velocity is 0, so there's no kinetic energy.

And the potential energy remains the same (i.e constant) at any given point in the system.

Using P.E = mgh.

At initial position, PE1 = mgh

At final position, PE2 = (m + M)gH.

Where h and H represent the initial and final heights.

m + M is the new weight after Jane and Tarzan swing

Equating PE1 to PE2

mgh = (m + M)gH

By substituton (M = 2m)

mgh = (m + 2m)gH

mgh = 3mgH

Make H the subject of the formula

H = mgh/3mg

H = ⅓h

Hence, the maximum height Tarzan and Jane can swing as a fraction of her initial heigh is ⅓h

From the question, the new height looks to be about ½ that of Jane's original position; i.e. ½h

The calculated height is smaller than what the cartoon is showing;

We can conclude that the cartoon is wrong.

4 0

3 years ago

If a quasar is moving away from us at v/c = 0.8, what is the measured redshift?

Delicious77 [7]

Answer:

The measured redshift is z =2

Explanation:

Since the object is traveling near light speed, since v/c = 0.8, then we have to use a redshift formula for relativistic speeds.

$z= \sqrt{\cfrac{c+v}{c-v}}-1$

Finding the redshift.

We can prepare the formula by dividing by lightspeed inside the square root to both numerator and denominator to get

$z= \sqrt{\cfrac{1+\cfrac vc}{1-\cfrac vc}}-1$

Replacing the given information

$z= \sqrt{\cfrac{1+0.8}{1-0.8}}-1$

$z= \sqrt{\cfrac{1.8}{0.2}}-1\\z= \sqrt{9}}-1\\z=3-1\\z=2$

Thus the measured redshift is z = 2.

4 0

3 years ago

Which is the currrent SI unit to use in measuring the amount of material contained in a solid sphere

olya-2409 [2.1K]

Hey

SI unit for measure the amount of material contained in a solid sphere metre cube(m^3).

mean it have Volume .

3 0

4 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago