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d1i1m1o1n [39]
3 years ago
15

How much force is required to accelerate a 5 kg mass at 20 m/s^2

Physics
1 answer:
Studentka2010 [4]3 years ago
6 0

Hello!

\large\boxed{F = 100N}

Use the equation F = m · a (Newton's Second Law) to solve. Substitute in the given values:

F = 5 · 20

F = 100N

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Heat pumps move heat from __________ to __________ areas & REQUIRES energy. *
sergejj [24]
The correct answer to this question is false
8 0
3 years ago
Date<br>Page<br>What are the advantages of alcohol<br>thermometric liquid?<br>​
Leya [2.2K]

Answer:

low freezing point. high vapour pressure.

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3 0
3 years ago
An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e
White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

5 0
3 years ago
Two tugboats are moving a barge. Tugboat A pushes on the barge with a force of 3000n. Tugboat B pulls with a force of 5000 Newto
kenny6666 [7]

The net force on the barge is 8000 N

Explanation:

In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.

In this problem, we have two forces:

  • The force of tugboat A, F_A = 3000 N, acting in a certain direction
  • The force of tugboat B, F_B = 5000 N, also acting in the same direction

Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get

F=F_A+F_B = 3000 + 5000 = 8000 N

and the direction is the same as the direction of the two forces.

Learn more about forces:

brainly.com/question/11179347

brainly.com/question/6268248

#LearnwithBrainly

5 0
4 years ago
What is the distance of separation between objects of masses 5.6 x 10 5 kg and 8.8 x 10 6 kg if the force of gravity between the
AlladinOne [14]

Answer:

2.87m

Explanation:

Using the law of gravitation to solve this question

F = GMm/r²

G is the gravitational constant

M and m are the masses

r is the distance between the masses

Substitute the given values

G = 6.67×10^-11 m³/kgs²

M =8.8 x 10^6 kg

m = 5.6 x 10^5 kg

F =440N

400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²

400r² = 328.698×10

400r² = 3286.98

r² = 3286.98/400

r² = 8.21745

r = √8.21745

r = 2.87m

Hence the distance of separation is 2.87m

7 0
3 years ago
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