In which direction will the object accelerate?

A car (mass of 830 kg) is sitting on a car lift in a shop (neglect the mass of the lift itself). While the car is being lifted u

OLEGan [10]

There are two force acting on an object that is being lifted. (1) the weight of the car, (2) the upward force. The difference of these force should be equal to the product of the mass and the acceleration. (This is the content of Newton's 2nd Law of Motion). If we let the lifting force be F,
F - (830)(9.8) = (830)(3.8)
The value of F from the equation is 11288 N.

3 0

3 years ago

If two cars are 5 m apart, and one car has a mass of 2,565 kg and the other

alekssr [168]

Answer:

D

Explanation:

just did

7 0

3 years ago

A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori

vladimir2022 [97]

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

The horizontal component of the pulling force, $F cos \theta$ , where F = 50 N is the magnitude and $\theta=60^{\circ}$ is the angle between the direction of the force and the horizontal; this force acts in the forward direction
The force of friction, $F_f$ , acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

$\sum F_x = ma_x$

where

m is the mass of the block

$a_x$ is the horizontal acceleration

However, the block is moving at constant speed, so the acceleration is zero:

$a_x = 0$

So the equation becomes

$\sum F_x = 0$ (1)

The net force here is given by

$\sum F_x = F cos \theta - F_f$ (2)

And so, by combining (1) and (2), we find the magnitude of the friction force:

$F cos \theta - F_f = 0\\F_f = F cos \theta = (50)(cos 60^{\circ})=25 N$

Learn more about force of friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

4 0

3 years ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, using the Parallel Axis Theorem, we calculate I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$

5 0

3 years ago

theres two cups on a board - one half full and one empty. which one will fall off quicker while lifting the board up?

34kurt

The half empty will fall quicker

Please leave a thanks <3

6 0

3 years ago