In the laboratory you will find three bottles with the following labels:

You can enter compound units that are combinations of other units that are multiplied together. to enter the ⋅ explicitly, type

alexdok [17]

Answer:

$30 N \cdot m$

Explanation:

The torque applied by a force can be calculated as

$\tau = F d sin \theta$

where

F is the magnitude of the force

d is the length of the arm

$\theta$ is the angle between the direction of the force and the arm

In this problem, we have

F = 15 N

d = 2.0 m

$\theta=90^{\circ}$

Substituting into the equation, we find

$\tau = (15)(2.0) sin 90^{\circ}=30 N \cdot m$

7 0

3 years ago

Please answer as fast as possible

aleksley [76]

Electric pumps are not useful they clog

6 0

2 years ago

Read 2 more answers

Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that ap

eduard

Answer:

The object can have zero velocity and, simultaneously, nonzero acceleration.

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

Explanation:

An object in simple harmonic motion has a total mechanical energy (sum of elastic potential energy and kinetic energy) that is constant:

E=U+K=1/2kx^2 + 1/2}mv^2

where,

k is equal to the spring constant

x is equal to the displacement

m is the mass

v is the speed

We can note that the force on the spring is given by Hook's law:

F=-kx

In Newton's law F = ma, this can be also be written as

ma=-kx

a=-k/mx

This implies that the acceleration is proportional to the displacement.

From the first equation, we can now states that:

When the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum

When the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

In all the other intermediate situations, both velocity and acceleration are nonzero.

So the correct answers are

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

The object can have zero velocity and, simultaneously, nonzero acceleration.

4 0

3 years ago

Read 2 more answers

A 30 kg child went down a 10 m tall slide. Assuming no energy was lost as friction, what was the child's velocity when he reache

AveGali [126]

14 m/s or 50km/h. See the details in the attached picture.

5 0

2 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

2 years ago