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3 years ago

7

a ball dropped from rest falls freely intil it hits the ground with the speed of 20 m/s . the tine furing which the ball is in f

ree fall is approximately

1 answer:

forsale [732]3 years ago

4 0

Answer:

5m/s

Explanation:

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If youre walking from point a to b, the magnitude of your displacement will always be equal or less than or greater than your di

xenn [34]

The magnitude of your displacement can be equal to the distance you covered, or it can be less than the distance you covered. But it can never be greater than the distance you covered.

This is because displacement is a straight line, whereas distance can be a straight line, a squiggly line, a zig-zag line, a line with loops in it, a line with a bunch of back-and-forths in it, or any other kind of line.

The straight line is always the shortest path between two points.

3 0

2 years ago

A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressin

BARSIC [14]

Answer:

$U_k = 0.113$

Explanation:

using the law of the conservation of energy:

$E_i -E_f=W_f$

$\frac{1}{2}Kx^2=NU_kd$

where K is the spring constant, x is the spring compression, N is the normal force of the block, $U_k$ is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

N = mg

N = 3.35 kg * 9.81 m/s

N = 32.8635

So, Replacing values on the first equation, we get:

$\frac{1}{2}(138)(0.123)^2= (32.8635)U_k(0.281m)$

solving for $U_k$ :

$U_k = 0.113$

8 0

3 years ago

1.which of the following are true.

Ksivusya [100]

B is the correct answer

3 0

3 years ago

Read 2 more answers

Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th

brilliants [131]

From the case we know that:

The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

#SPJ4

8 0

1 year ago

An open organ pipe 30 cm long and a closed organ pipe 23 cm long, both of same diameter , are each sounding its first overtone ,

cupoosta [38]

First overtone of open organ pipe is given as

$f_{1o} = \frac{v}{L_1 + 2e}$

first overtone of closed organ pipe is given as

$f_{1c} = \frac{3v}{4(L_2 + e)}$

now they are in unison so we will have

$\frac{v}{L_1 + 2e} = \frac{3v}{4(L_2 + e)}$

$\frac{1}{30 + 2e} = \frac{3}{4(23 + e)}$

$90 + 6e = 92 + 4e$

$e = 1 cm$

so end correction of both pipes is e = 1 cm

8 0

3 years ago