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2 years ago

3

1. A 5.00 kg box is being pulled at constant speed up a 30o incline by a force of 45.0 N parallel to the incline. Determine the

coefficient of kinetic friction between the box and the plane.

1 answer:

tangare [24]2 years ago

6 0

Answer:

μ = 0.48

Explanation:

To solve this problem we must make a free body diagram locating each of the respective forces on the boxes.

We must pay special attention that the body is moving at a constant speed, therefore there is no acceleration, and the sum of forces is equal to zero.

Attached is an image with the sum of forces and the equations of the sum of forces on the x & y axes.

The solution of all equations can be seen, in the attached image.

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Vectors a and b have scalar product â6.00, and their vector product has magnitude +9.00. what is the angle between these two vec

salantis [7]

Answer:

Value of angle between vector a and b is $56.30^{\circ}$ .

Explanation:

Vectors a and b have scalar product 6.00

Let $\theta$ be the angle between a and b.

$\vec{a}.\vec{b} = 6$

ab cos $\theta$ = 6 ...(1)

Vectors a and b have magnitude of vector product 9.00

$\vec{a} \times\vec{b} = 9$

ab sin $\theta$ = 9 ...(2)

Dividing equation (2) by (1) we get

$\frac{ab sin \theta}{ab cos \theta} = \frac{9}{6}$

tan $\theta$ = 1.5

$\theta = tan ^{-1} (1.5)$

$\theta$ = $56.30^{\circ}$

Thus, value of angle between vector a and b is $56.30^{\circ}$ .

3 0

2 years ago

Consider two slides, both of the same height. One is long and the other is short. From which slide will a child have a greater f

Lunna [17]

Answer:

The final speed will be the same for the children on the shorter side and on the longer side.

Explanation:

This is because since the they are the same distance above the ground, their potential energy which is a function of mass, acceleration due to gravity and vertical height are the same.

PE = Mass × gravity × vertical height

At this point, we can deduce that the horizontal length of the slide has no effect on the potential energy. Only the vertical height does.

All this potential energy is converted to kinetic energy at the end of the slide. Since the potential energy is the same, then the kinetic energy will be the same and thus their velocity is the same.

Mathematically, consider that PE = mgh and KE = $\frac{1}{2}$ m $v^{2}$

at the bottom of the slide, since energy has to be conserved, PE must be equal to KE.

mgh = $\frac{1}{2}$ m $v^{2}$

final velocity of the child , v = $\sqrt{2gh}$

It shows the final velocity is only a function f acceleration due to gravity and height.

Thus, making their velocities equal.

8 0

2 years ago

A 2-kg wood block is pulled by a string across a rough horizontal floor. The string exerts a tension force of 30 N on the block

Luden [163]

Answer:

Work done, W = 84.57 Joules

Explanation:

It is given that,

Mass of the wooden block, m = 2 kg

Tension force acting on the string, F = 30 N

Angle made by the block with the horizontal, $\theta=20^{\circ}$

Distance covered by the block, d = 3 m

Let W is the work done by the tension force. It can be calculated as :

$W=F\ cos\theta\times d$

$W=30\times cos(20)\times 3$

W = 84.57 Joules

So, the work done by the tension force is 84.57 Joules. Hence, this is the required solution.

7 0

2 years ago

what velocity must a 1340kg car have in order to havw the same momentum as a 2680 kg truck traveling at a velocity of 15m/s to t

kykrilka [37]

Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.

Hope i helped
Have a good day :)

6 0

2 years ago

A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

Mice21 [21]

Answer:

$v = 3369.2 m/s$

Explanation:

As we know that Beacon is rotating with angular speed

$f = 10 rev/min$

so we have

$\omega = 2\pi f$

$\omega = 2\pi(\frac{10}{60})$

$\omega = 1.047 rad/s$

now we know that

$v = r \omega$

here we will have

$r = 2 miles = 2(1609 m)$

$r = 3218 m$

so we have

$v = 3218(1.047)$

$v = 3369.2 m/s$

6 0

2 years ago