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krek1111 [17]
3 years ago
9

When a 4.60-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.30 cm. (a) If

the 4.60-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it? cm (b) How much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position?

Physics
1 answer:
Alex787 [66]3 years ago
6 0

Answer:

a) y = 0.0075 m

b) W = 1.569 J

Explanation:

See attachment for the solution

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A ball initially at rest rolls down a hill with an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move?
goldfiish [28.3K]

<u>Answer:</u>

  Ball will move 92.8125 meter along the cliff in 7.5 seconds.

<u>Explanation:</u>

We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case initial velocity = 0 m/s, acceleration = 3.3 m/s^2, we need to calculate displacement when time = 7.5 seconds.

Substituting

  s=0*7.5+\frac{1}{2} *3.3*7.5^2\\ \\ =92.8125 meter

  So ball will move 92.8125 meter along the cliff in 7.5 seconds.

5 0
3 years ago
Amanda spent 2/5 of her time after school doing homework and ¼ of her remaining time riding her bike. If she rode her bike for 4
Doss [256]

Answer:120 min

Explanation:

Given

Amanda  spent \frac{2}{5} of her time after school doing Home work

And \frac{1}{4} of her remaining  time riding her bike

It is given that she rode her bike for 45 minutes in a week

Let t be the time after school

therefore Amanda spend \frac{2t}{5} in home work and  \frac{3t}{5} time is left

From remaining \frac{3t}{5} time she spends \frac{1}{4} time riding her bike

therefore \frac{3t}{5}\times \frac{1}{4}=45

thus t=300 min

therefore time  spent on home work is \frac{2}{5}\times 300=120 min

6 0
3 years ago
Which of these boxes will not accelerate? A. B. C. D.
serg [7]

Answer: The correct answer is Image B.

Explanation: For an object to accelerate, there should be unbalanced forces present. An object will move in the direction of net force.

Balanced forces are defined as the forces acting on the same object which are equal in magnitude but act in opposite direction. The net forces are 0.

Unbalanced forces are defined as the forces acting on the same object which are unequal in magnitude. The net force is non-zero.

For the given images:

Image A: This box will accelerate easily because the net force is non-zero and is moving in right direction.

Image B: This box will not accelerate because the net force is zero as all the forces are balancing one another. Hence, the object will stay at rest.

Image C: This box will accelerate easily because the net force is non-zero and  is acting in between the normal and applied force.

Image D: This box will accelerate easily because the net force is non-zero and is moving in right direction.

Hence, the correct option is Image B.

7 0
3 years ago
Read 2 more answers
A small sculpture made of brass (rho brass = 8470 kg/m 3 ) is believed to have a secret central cavity. The weight of the sculpt
GaryK [48]

Answer:

Volume of secret cavity = 4 x 10⁻⁶m³

Explanation:

The weight of the sculpture in air is 15.76 N

Mass of the sculpture = 1.61 kg

Mass = Volume x Density

1.61 = V x 8470

Volume of brass =1.90 x 10⁻⁴ m³

When it is submerged in water, the weight is 13.86 N.

That is

  Weight of sculpture - Weight of water displaced = 13.86 N

  15.76 - Weight of water displaced = 13.86

   Weight of water displaced = 1.9 N

   Mass of water displaced = 0.194 kg

 Mass = Volume x Density

  0.194 = V x 1000

  Volume of water displaced =1.94 x 10⁻⁴ m³

Volume of secret cavity =  Volume of water displaced - Volume of brass material

Volume of secret cavity = 1.94 x 10⁻⁴-1.94 x 10⁻⁴ = 0.04x 10⁻⁴ = 4 x 10⁻⁶m³

7 0
3 years ago
A 50 kg archer standing on frictionless ice shoots a 100g arrow at a speed of 100 m/s what uis the recoild pseed of the archer?
nexus9112 [7]

Answer:

v= 0.2 m/s

Explanation:

Given that

m₁ = 50 kg

m₂ = 100 g = 0.1 kg

u =10 0 m/s

If there is no any external force on the system  then the total linear momentum of the system will be conserve.

Initial linear momentum = Final momentum

m₁u₁ + m₂ u₂ =m₂ v₂ +m₁v₁

50 x 0 + 0.1 x 100 = 50 v + 0

0+ 10 = 50 v

v=\dfrac{10}{50}\ m/s

v= 0.2 m/s

Therefore the recoil speed will be 0.2 m/s.

7 0
3 years ago
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