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miss Akunina [59]
3 years ago
15

If the time of impact in a collision is extended by 3 times, by hour much is

Physics
1 answer:
vodomira [7]3 years ago
4 0

Answer:

The force of the impact would be smaller

Explanation:                                                          Examples:

If the force is big then the time would be small (2500N of Force = 10 seconds)

If the force is small then the time would be big (250N of Force = 50 seconds)

Impulse/Collision -> [Ft] = [M (vf-vo)] <- Change in momentum

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If a 0.15 kg ball falls and has a KE of 20 J just before striking the ground, from what height did it fall. A. 1.36m B. 3m C. 13
RUDIKE [14]
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell. 

Therefore, we can say KE = PE
We know that PE = m·g·h

Which means KE = m·g·h

We can solve for h:

h = KE / m·g
   = 20 / (0.15 · 9.8) 
   = 13.6m

The correct answer is: the ball has fallen from a height of 13.6m.

5 0
3 years ago
If one of two interacting charges is doubled, the force between the charges will _____________.
malfutka [58]

If one of two interacting charges is doubled, the force between the charges will double.

Explanation:

The force between two charges is given by Coulomb's law

F=\frac{k q1 q2}{r^{2}}

K=constant= 9 x 10⁹ N m²/C²

q1= charge on first particle

q2= charge on second particle

r= distance between the two charges

Now if the first charge is doubled,

we get F'=\frac{k (2q1) q2}{r^{2}}

F'= 2 F

Thus the force gets doubled.

4 0
3 years ago
In order for work to take place the energy present must be related to the movement of the object.
kaheart [24]

This is another one of those muddy misleading questions, followed by
a muddy group of choices from which an answer must be selected.

a).  is absurd.  There's no such thing as a "balanced force", only
a balanced group of forces.

b).  is probably the choice the question is aiming for.

c).  is not so.  The engines of an airplane do plenty of work lifting the plane
off the ground, although the force of the engines is never directed upward.

d).  is really awkward.  The object's motion is almost never the cause of the force.
The force is almost always the cause of the object's motion.

Now for the big 800-lb gorilla in the room:  No moving object needs to be involved
in order for energy to be flowing or work to be getting done.

-- A radio wave radiates through space.  Straighten out a wire coat-hanger and
stick it up in the air where the radio wave can pass by it.  Electrical current flows
through the wire, and you can drain the electrical energy out the bottom of it.  

-- A light bulb is shining.  Some distance away, something it's shining on
gets warm, because of the heat energy that has shot across to it from the
light bulb and soaked into it.

-- A lightning bolt jumps from the ground to a passing cloud.  Or, if you feel
more comfortable with it, a lightning bolt jumps from a cloud to the ground.
It doesn't matter.  Either way, there's enough energy splashing around to
ignite houses, zap TVs and computers, melt concrete, vaporize water, and
light up a city.  Although nothing is moving.

5 0
3 years ago
A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang
Dafna11 [192]

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) E=\frac{1}{2}mv^2+mgh

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) E=\frac{1}{2}mv^2

The energy when the mass comes to a stop will be:

3) E=mgh

Setting equations 2 and 3 equal and solving for height h will give:

4) h=\frac{v^2}{2g}

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) cos\phi=\frac{l-h}{l}

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) \phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})

8 0
3 years ago
Use Hooke's Law to determine the variable force in the spring problem. A force of 450 newtons stretches a spring 30 centimeters.
ladessa [460]

Answer:

Work Done = 67.5 J

Explanation:

First we find the value of spring constant (k) using Hooke's Law. Hooke's is formulated as:

F = kx

where,

F = Force Applied = 450 N

k = Spring Constant = ?

x = Stretched Length = 30 cm = 0.3 m

Therefore,

450 N = k(0.3 m)

k = 450 N/0.3 m

k = 1500 N/m

Now, the formula for the work done in stretching the spring is given as:

W = (1/2)kx²

Where,

W = Work done = ?

k = 1500 N/m

x = 70 cm - 40 cm = 0.3 m

Therefore,

W = (1/2)(1500 N/m)(0.3 m)²

<u>W = 67.5 J</u>

3 0
3 years ago
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