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3 years ago

If we decrease the distance an object moves we will

2 answers:

6 0

... without changing the speed of the object, then
we will not have to wait so long for the object to get
where it is going.

aleksley [76]3 years ago

4 0

If we decrease the distance an object moves, we will decrease the amount of work done.

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You have been hired to help improve the material movement system at a manufacturing plant. Boxes containing 16 kg of tomato sauc

pickupchik [31]

a) $15.4^{\circ}$

b) 5.2 m/s

c) 151.2 N

Explanation:

When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by

$mg sin \theta$

where

m =16 kg is the mass of the box

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

$\theta$ is the angle of the ramp

According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:

$F=ma\\mgsin \theta = ma$

where a is the acceleration.

From the equation above we get

$a=g sin \theta$

And we are told that the acceleration must not exceed

$a=2.6 m/s^2$

Substituting this value and solving for $\theta$ , we find the maximum angle of the ramp:

$\theta=sin^{-1}(\frac{a}{g})=sin^{-1}(\frac{2.6}{9.8})=15.4^{\circ}$

Here we are told that the vertical distance of the ramp is

$h=1.4 m$

Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.

So we have:

$GPE=KE\\mgh=\frac{1}{2}mv^2$

where:

m = 16 kg is the mass of the box

$g=9.8 m/s^2$

h = 1.4 m height of the ramp

v = final speed of the box at the bottom of the ramp

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.4)}=5.2 m/s$

There are two forces acting on the box in the direction perpendicular to the ramp:

- The normal force, N, upward

- The component of the weight perpendicular to the ramp, downward, of magnitude

$mg cos \theta$

Since the box is in equilibrium along the perpendicular direction, the net force is zero, so we can write:

$N-mg cos \theta$

and by substituting:

m = 16 kg

$g=9.8 m/s^2$

$\theta=15.4^{\circ}$

We can find the normal force:

$N=mg cos \theta=(16)(9.8)cos(15.4^{\circ})=151.2 N$

8 0

3 years ago

Which of the statements below is true?

Cloud [144]

Answer:

An object’s weight is proportional to its mass.

8 0

3 years ago

A racecar is equipped with a computer that records the reading on its speedometer every second during a race. If you graph this

Tanya [424]

Because it records speed of the car at a certain time, the independent variable should be time and dependent would be speed or velocity. Since it's taken every second, it would be considered instantaneous velocity, which is D.

4 0

3 years ago

The planet Jupiter has much more mass than Earth. What would happen to an astronaut who landed on Jupiter

Ivanshal [37]

Answer:

A. The astronaut would way somewhat less

5 0

3 years ago

Read 2 more answers

Como calcular massa volume e densidade?

shepuryov [24]

Mass divided by volume
Piensa así M
--- = D
V
Como un corazón
Masa división por volumen igual a densidad