The table below shows the mass and velocity of four objects. Which object has the least inertia?A. Y
B. Z
C. X
D. W

Answers

Answer 1

Answer:

The object with the least inertia is Z.

option B is the correct answer.

What is Newton's first law of motion?

Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that path unless it is acted upon by an external force and it will move in the direction of applied force.

The Newton's first law of motion is also called the law of inertia because it depends on the mass of the object.

Inertia is defined as the reluctancy of an object to move when a force is applied to it.

As the mass of an object increase, the inertia of the object increases because the object will be more reluctant to move when a force is applied to it.

Thus, the more massive an object is, the greater the object's inertia and vice versa.

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The graph below shows the velocity of a car as it attempts to set a speedrecord.Velocity vs. Time140013001200110010004 4531 (s)At what point is the car the fastest?A. t = 1.0 sB. t = 4.2 sC. t = 3.0 sD. t = 4.5 s
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25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression in terms of the variables d, vcmax and vg for the time, tc, it takes the cheetah to catch the gazelle.

Answers

maximum speed of cheetah is

$v_1 = v_(max)$

speed of gazelle is given as

$v_2 = v_(g)$

Now the relative speed of Cheetah with respect to Gazelle

$v_(12) = v_1 - v_2$

$v_(12) = v_(max) - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_(12)* t$

so by rearranging the terms we can say

$t = (d)/(v_(12))$

$t = (d)/(v_(max) - v_g)$

so above is the relation between all given variable

Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done by the force

Answers

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

$\Rightarrow a=(F)/(m)$

Hence acceleration of the block is given by:

$\Rightarrow a=(25)/(2)=12.5m/s^2$

Displacement of the object is given by:

$\Rightarrow S=ut+(1)/(2)at^2$

Substitute the values

$\Rightarrow S=0*3+(1)/(2)(12.5)3^2$

$\Rightarrow S=56.25 m$

Now work done is given by:

W=F.S

W = 25×56.25

W =1406.25 J

Dawn is trying to find out how much weight she can push across the room. She is really trying to find her

Answers

Dawn is trying to figure out how much weight she can push with her strength, or what her maximum pushing force is, across the room. She could do an experiment to find out.

She must first prepare a testing space with a flat, smooth surface to reduce friction. She can then progressively add weights to a cart or other object and use all of her strength to try to push it across the room. She can determine her maximum pushing force by noting the heaviest weight she can move. For a variety of jobs, including moving furniture or participating in physical sports that call for pushing heavy things, this knowledge can be essential.

To know more about physical sports, here

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Answer:

Muscular strength

Explanation:

She is testing her strength while pushing the weights

22. What force is necessary to accelerate a 2500kg car from rest to 20m/s over 10s?(6 Points)
2N
250N
5000N
50000N

Answers

Answer:

50000N

Explanation:

Force = mass × acceleration

= 2500 × 20

= 50000N

A hollow sphere of radius 0.25 m is rotating at 13 rad/s about an axis that passes through its center. the mass of the sphere is 3.8 kg. assuming a constant net torque is applied to the sphere, how much work is required to bring the sphere to a stop?

Answers

The work required to bring the sphere to stop is equal to the kinetic energy possessed by the sphere.

Kinetic energy of a rotating body is given by,

K.E = $(1)/(2)Iw^(2)$

Here, I= Moment of inertia of hollow sphere,

Since, the hollow sphere is rotating about the axis passing through its center, I = $(2)/(3)MR^(2)$

M= Mass of the sphere= 3.8 kg,

R= Radius of gyration= Radius of the sphere= 0.25 m

w= Angular speed of the sphere = 13 rad/s

Substituting the values,

Kinetic energy = $(1)/(2) *(2)/(3) (3.8)(0.25)^(2)(13.0)^(2)$

= 13.4 J

∴ Work required to bring the sphere to stop is 13.4 J.

The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)