What class lever is a chest pass in netball?

A man pushes a shopping cart across a level floor. What force resists the effort force? A) gravity B) friction C) the normal for

Kipish [7]

Answer:

B) Friction

Explanation:

Friction is a force that acts when an object is sliding along a surface. Microscopically, this force is due to the fact that the two surfaces are not perfectly smooth, but they have "imperfections" that cause a force that opposes the motion of the object.

For an object sliding on a flat surface, the force of friction has magnitude:

$F_f = \mu_k mg$

where

$\mu_k$ is the coefficient of kinetic friction

m is the mass of the object

g is the acceleration of gravity

The direction of the force of friction is always opposite to the direction of motion of the object.

In reality, friction also acts if the object is at rest and it is pushed by a force; in this case, we talk about static friction, and its magnitude is

$F_f = \mu_s mg$

where $\mu_s$ is called coefficient of static friction, and it is generally larger than the coefficient of kinetic friction.

8 0

3 years ago

How much have jellyfish populated since 2015?

My name is Ann [436]

About a mil sience 2014-2015

4 0

3 years ago

Read 2 more answers

A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

$I_1\omega_1=I_2\omega_2$

Here, $I_2=I_1+mr^2$

$I_1\omega_1=(I_1+mr^2)\omega_2$

$900\times 0.95=(900+70\times (2.9)^2)\omega_2$

Solving the above equation, we get the value as :

$\omega_2=0.574\ rad/s$

Part B,

The initial rotational kinetic energy is given by :

$k_i=\dfrac{1}{2}I_1\omega_1^2$

$k_i=\dfrac{1}{2}\times 900\times (0.95)^2$

$k_i=406.12\ rad/s$

The final rotational kinetic energy is given by :

$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

$k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2$

$k_f=245.24\ rad/s$

Hence, this is the required solution.

5 0

3 years ago

Why is earth the only planet on which water exists in a liquid state

Gennadij [26K]

This is due to earths location in the solar system. Earth is in the habitat zone or the Goldie locks zone, in this zone it's not too hot or not too cold for water to exist. Other planets in different star systems have liquid oceans due to them being in the habitat zone.

7 0

3 years ago

What is the average velocity of the object from t=1s to t= 3 s?

galina1969 [7]

Answer:

Explanation:

Since this is a distance v time graph, the slope of the line from 1s to 3s is the velocity. However, it looks like, at t=3, the velocity is 0, so getting the definite velocity is not going to happen. We can estimate it as closely as possible. Since the line is tending from the upper left to the lower right, the slope is negative, so the velocity is also negative. That leaves only C or D as our answers. And the slope is closer to -1 than to -5, so choice D. is the one you want.

7 0

3 years ago