What is the resultant force acting on an object with an eastward force of

How to find aceleration

lions [1.4K]

Answer:

$a=\frac{v_{f}-v_{i}}{t_{f}-t_{i}}$

Explanation:

Formula for AVERAGE acceleration

$a=\frac{v_{f}-v_{i}}{t_{f}-t_{i}}$

$v_{f}$ is Final Velocity

$v_{i}$ is Initial Velocity

$t_{f}$ is Final Time

$t_{i}$ is Initial Time

Most of the time, the initial time will be 0, so you won't have to subtract.

5 0

3 years ago

A high-jumper clears the bar and has a downward velocity of - 5.00 m/s just before landing on an air mattress and bouncing up at

Jobisdone [24]

-- As she lands on the air mattress, her momentum is (m v)

Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down

-- As she leaves it after the bounce,

Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up

-- The impulse (change in momentum) is

Change = (60 kg-m/s up) - (300 kg-m/s down)

Magnitude of the change = 360 km-m/s

The direction of the change is up /\ .

8 0

3 years ago

A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the

sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

$x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg$

Since the frictional force $f_k$ is equal to $f_k=\mu_k N=\mu_k mg$ , then we have that the acceleration is:

$a=\frac{\mu_k mg}{m}=\mu_k g$

Now, from the definition of acceleration we get:

$a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}$

And, as the final velocity is zero because the box gets to a stop, we have:

$t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}$

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

$t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s$

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

$v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}$

Finally, we calculate the displacement:

$x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm$

This means that the box moves 7.29cm backwards relative to the belt.

4 0

3 years ago

Which of the following is a good scientific question? A. Are dragons lizards or birds? B. How do star fish fall in love? C. Does

Wittaler [7]

C. Does eating less fat increase a mouse's lifespan?
(If a mouse eats less fat in its meals then will it live longer: compared to other mice lives.

This is a testable question that can be answered by designing and conducting an experiment.

I think D could be correct as well.

My final answer is D.

6 0

3 years ago

Read 2 more answers

Two students make the following claims:

antiseptic1488 [7]

Answer:

E. Student 1 is correct, because as θ is increased, h is the same.

Explanation:

Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.

Mathematically:

$F=m.g$

As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.

And for work we have:

$W=F.s\cos\theta$

where:

$s=$ displacement of the object

$\theta=$ angle between the force and displacement vectors

Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.

In case of free-fall the angle between the force is and the displacement is zero.
In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.

So, the work done by the gravitational force is same in the two cases.

6 0

3 years ago