Ask question

Ask question

All categories

3 years ago

5

A cart is moving to the right with a constant speed of 20 m s . A box of mass 80 kg moves with the cart without slipping. The co

efficient of static friction between the box and cart is 0.30, and the coefficient of kinetic friction between the box and cart is 0.15. What are the magnitude and direction of the frictional force acting on the box as it moves with the cart?

1 answer:

nignag [31]3 years ago

4 0

Answer:

Explanation:

Given

mass of box $m=80 kg$

coefficient of kinetic friction $\mu _k=0.15$

coefficient of Static friction $\mu _s=0.30$

cart is moving with constant velocity therefore Net Force is zero

Since there is no net acceleration therefore friction force will be zero

mathematically

$f_r=ma$

where $f_r=frictional\ Force$

$a=acceleration$

$a=0$

$f_r=0$

You might be interested in

Which statement best describes the difference between atoms and molecules

Rama09 [41]

What are the answers ?

6 0

2 years ago

Read 2 more answers

Balancing equations<br>

kolbaska11 [484]

Answer:

Tips for Balancing Chemistry Equations You may only put numbers in front of the molecules. You can never alter the subscript, as it would alter the formula. ... Begin by counting and making note of the elements present on each side of the equation. Count the number of molecules of each element on both sides of the equation. More items...

Explanation:

I dont know what the question is

3 0

2 years ago

For a short time the missile moves along the parabolic path y=(18−2x2) km. If motion along the ground is measured as x=(4t−3) km

muminat

Answer with Explanation:

We are given that

$y=(18-2x^2) km$

$x=(4t-3)km$

Differentiate x and y w.r.t t

$\frac{dx}{dt}=4$

$\frac{dy}{dt}=-4x\frac{dx}{dt}$

$\frac{dy}{dt}=-4x\times 4=-16x=-16(4t-3)$

$v_x=\frac{dx}{dt}=4$

$v_y=\frac{dy}{dt}=-16(4t-3)$

Substitute t=1

$v_x=4$

$v_y=-16(4-3)=-16$

Magnitude of velocity= $\mid v\mid=\sqrt{v^2_x+v^2_y}$

$\mid v\mid=\sqrt{4^2+(-16)^2}=16.49 m/s$

Hence, the magnitude of the missile's velocity=16.49 m/s

$a_x=\frac{d(\frac{dx}{dt})}{dt}=\frac{d(4)}{dt}=0$

$a_y=\frac{d(\frac{dy}{dt})}{dt}=\frac{d(-16(4t-3))}{dt}=-64$

Substitute t=1

$a_x=0,a_y=-64$

$\mid a\mid=\sqrt{a^2_x+a^2_y}$

$\mid a\mid=\sqrt{0+(-64)^2}=64m/s^2$

Hence, the magnitude of acceleration when t=1 s= $64m/s^2$

3 0

2 years ago

HELP HELP HELP HELP HELPPPPPPPPPPP

Firdavs [7]

Initial velocity (u) = 2 m/s

Acceleration (a) = 10 m/s^2

Time taken (t) = 4 s

Let the final velocity be v.

By using the equation,

v = u + at, we get

or, v = 2 + 10 × 4

or, v = 2 + 40

or, v = 42

The final velocity is 42 m/s.

5 0

2 years ago

Read 2 more answers

You are playing a speed-based card game with your 64-year-old grandfather. The object of the game is to get rid of your cards as

Masteriza [31]

Now, you always beat him. Your grandfather is likely experiencing a slight decline in perceptual speed.

<u>Explanation:</u>

The speed of perception refers to the capacity to accurately (and completely) compare words letter, digits, objects, images, etc. When testing, these objects can be displayed simultaneously or one after the other. This type of test can be included in the proficiency test.

For example, we have also seen all the puzzles that ask the reader to notice the differences between the two pictures. The time it takes to recognize these differences is a measure of the speed of perception. Likewise, in getting rid of cards at the given situation, grandfather experiences a less decline in his perceptual speed.

4 0

3 years ago