All categories

3 years ago

Identify the type of friction in each situation described below:

Physics

1 answer:

-BARSIC- [3]3 years ago

8 0

Answer:

A . static frictional force

B. kinectic friction

C.. Rolling friction

You might be interested in

A 225 kg block is pulled by two horizontal forces. The first force is 178 N at a 41.7-degree angle and the second is 259 N at a

yawa3891 [41]

Answer:

52.9 N, 364.7 N

Explanation:

First of all, we need to resolve both forces along the x- and y- direction. We have:

- Force A (178 N)

$A_x = (178 N)(cos 41.7^{\circ})=132.9 N\\A_y = (178 N)(sin 41.7^{\circ})=118.4 N$

- Force B (259 N)

$B_x = (259 N)(cos 108^{\circ})=-80.0 N\\B_y = (259 N)(sin 108^{\circ})=246.3 N$

So the x- and y- component of the total force acting on the block are:

$R_x = A_x + B_x = 132.9 N - 80.0 N =52.9 N\\R_y = A_y + B_y = 118.4 N +246.3 N = 364.7 N$

7 0

3 years ago

If 20 beats are produced within a single second, which of the following frequencies could possibly be held by two sound waves tr

zhuklara [117]

The correct choice is

D. 22 Hz and 42 Hz.

In fact, the beat frequency is given by the difference between the frequencies of the two waves:

$f_B = |f_1 -f_2|$

In this problem, the beat frequency is $f_B=20 Hz$ , therefore the only pair of frequencies that gives a difference equal to 20 Hz is

D. 22 Hz and 42 Hz.

4 0

3 years ago

Read 2 more answers

What foods are recommended for pregnant woman?

miss Akunina [59]

Answer:

Vagetagble, fruits,fresh milk and juice vitamins c

3 0

3 years ago

How are theories and laws connected

Anni [7]

Answer:

Laws are statements about something that's been observed and stated while a theory is an explanation of what's been observed. This connection between them forms a main idea that many people regulate as "what's normal."

Explanation:

4 0

2 years ago

Read 2 more answers

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago