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Svetlanka [38]
2 years ago
12

A body of mass 3.0Kg is acted upon by a force of 24N, if the frictional force on the body is 13N.Calculate the acceleration of t

he body​
Physics
1 answer:
lisabon 2012 [21]2 years ago
4 0

Answer:

Fnet=ma

24-13=3a

11/3=a

a=3.6m/s2

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Ever tried to stop a 150-pound (68 kg) cannonball fired towards you at 30 mph (48 km/hr.)? No, probably not. But you may have tr
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The two situations are similar because in both you are trying to minimize the damage and make the best out of a bad situation
8 0
2 years ago
A ping-pong ball weighs 0.025 N. The ball is placed inside a cup that sits on top of a vertical spring. If the spring is compres
kondor19780726 [428]

Answer:

Explanation:

The energy stored in the spring is used to throw the ball upwards . Let the height reached be h

stored energy of spring = 1/2 k y² , k is spring constant and y is compression created in the spring

stored energy of spring = potential energy of the ball

1/2 k y² = mgh , m is mass of the ball , h is height attained by ball

.5 k x .055² = .025  x 2.84

.0015125 k = .071

k = .071 / .0015125

= 46.9 N / m .

4 0
2 years ago
Use the momentum equation for photons found in this week's notes, the wavelength you found in #3, and Plank’s constant (6.63E-34
Nostrana [21]
To help you I need to assume a wavelength and then calculate the momentum.

The momentum equation for photons is:

p = h / λ , this is the division of Plank's constant by the wavelength.

Assuming λ = 656 nm = 656 * 10 ^ - 9 m, which is the wavelength calcuated in a previous problem, you get:

p = (6.63 * 10 ^-34 ) / (656 * 10 ^ -9) kg * m/s

p = 1.01067 * 10^ - 27 kg*m/s which  must be rounded to three significant figures.

With that, p = 1.01 * 10 ^ -27 kg*m/s

The answers are rounded to only 2 significan figures, so our number rounded to 2 significan figures is 1.0 * 10 ^ - 27 kg*m/s

So, assuming the wavelength λ = 656 nm, the answer is the first option: 1.0*10^-27 kg*m/s.
7 0
2 years ago
Read 2 more answers
Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the b
Salsk061 [2.6K]

Answer:

Explanation:

GIVEN DATA:

Distance between keisha and her friend 8.3 m

angle made by keisha toside building 30 degree

height of her friend monique is 1.5 m

from the figure

\Delta ACB

tan 30 = \frac{8.3}{h}

h= \frac{8.3}{tan 30} = 14.376 m

therefore

height of keisha is = h  + 1.5 m

                               = 14.376 + 1.5

= 15.876 \simeq 16 m

therefore option c is correct

5 0
2 years ago
With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by
Stells [14]
For Newton's second law, the resultant of the forces acting on the book is equal to the product between the mass of the book and its acceleration:
\sum F = ma (1)

There are only two forces acting on the book:
- its weight, directed downward: mg
- the force exerted by the hand on the book, of 20 N, directed upward

so, equation (1) becomes
mg - F = ma
from which we can calculate the book's acceleration, a:
a= g -  \frac{F}{m}= 9.81 m/s^2 - \frac{20 N}{3 kg}=3.14 m/s^2
7 0
2 years ago
Read 2 more answers
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