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ICE Princess25 [194]
2 years ago
12

Need the answer for question 5 :)

Physics
1 answer:
kykrilka [37]2 years ago
3 0

Answer:

1 B. Convert v from km/min to m/s ( show work and units

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What is an elastic collision?
Anit [1.1K]

Answer:

A collision in which both total momentum and total kinetic energy are conserved

Explanation:

In classical physics, we have two types of collisions:

- Elastic collision: elastic collision is a collision in which both the total momentum of the objects involved and the total kinetic energy of the objects involved are conserved

- Inelastic collision: in an inelastic collision, the total momentum of the objects involved is conserved, while the total kinetic energy is not. In this type of collisions, part of the total kinetic energy is converted into heat or other forms of energy due to the presence of frictional forces. When the objects stick together after the collision, the collisions is called 'perfectly inelastic collision'

6 0
3 years ago
Estimate the constant rate of withdrawal (in m3 /s) from a 1375 ha reservoir in a month of 30 days during which the reservoir le
kap26 [50]

Answer:

Explanation:

1 ha = 10⁴ m²

1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²

In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³

Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m

Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³

Let Q be the withdrawal in m³

Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶

Q = 26.20 x 10⁶ m³

rate of withdrawal per second

= 26.20 x 10⁶ / 30 x 24 x 60 x 60

= 26.20 x 10⁶ / 2.592 x 10⁶

= 10.11 m³ / s

6 0
3 years ago
A person's _____________ will change if they move from the Earth to the moon.
Diano4ka-milaya [45]

1. A person's weight will change if they move from the Earth to the moon.

In fact, the weight of a person is given by:

W=mg

where m is the mass of the person and g is the gravitational acceleration. The mass of the person, m, is the same on the Earth and on the moon, but the value of g is different on the Moon (about 1/6 of the Earth's value), so the weight also changes.


2. An astronaut is launched into space. The mass of the astronaut did not change. This is measured in Kg.

The mass of an object (or of a person, as in this case) is an intrinsec property of the object, that depends on the amount of matter inside the object: therefore, this quantity does not depend on the location of the object, so it is the same on the Earth, on the Moon and in space.


3. What is the weight of a ring tailed lemur that has a mass of 10 kg? -98 N

The weight of the lemur is given by:

W=mg

where m=10 kg is the lemur's mass and g=-9.8 m/s^2 is the gravitational acceleration. Using these numbers, we find

W=(10 kg)(-9.8 m/s^2)=-98 N

and the negative sign simply means that the direction of the weight is downward.


4. What is the mass of the lemur from the previous question if it was on the International Space Station? 10 kg

As we said in question 1), the mass of an object does not depend on the location, so the mass of the lemur is still 10 kg, as in the previous exercise.


5. A rocket being thrust upward as the force of the fuel being burned pushes downward is an example of which of Newton's laws? Third's Newton Law

Third's Newton Law states that:

"When an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A".

Applied to this case, the two objects are the fuel and the rocket. The fuel is pushed backward by the rocket, so the fuel exerts an equal and opposite force on the rocket, which then moves forward.


6. When a cannon is fired, the projectile moves forward. According to Newton's 3rd law, the cannon will want to travel backward.

Third's Newton Law states that:

"When an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A".

Applied to this case, the two objects are the cannon and the projectile.The projectile is pushed forward by the cannon, so the projectile exerts an equal and opposite force on the cannon, which moves backward.


7. An object has a weight of 21,532 N on Earth. What is the mass of the object? 2,197 kg

The weight of the object is given by: W=mg

If we re-arrange the formula and we use W=21,532 N, we can find the mass of the object:

m=\frac{W}{g}=\frac{21,532 N}{9.8 m/s^2}=2,197 kg


8. What is the mass of the object from the previous question if we put it on the moon? The force of gravity on the moon is 1.62 m/s2.  2,197 kg

As we said in question 4), the mass of an object does not change if we move it to another location, so its mass is still 2,197 kg.


9. How much force is exerted if a 250 kg object has an acceleration of 750 m/s2 ? 187,500 N

The force exerted on the object is given by Newton's second law:

F=ma

where F is the force, m=250 kg is the mass and a=750 m/s^2 is the acceleration. By using these numbers, we find

F=(250 kg)(750 m/s^2)=187,500 N


10. A resting soccer ball moving after it is kicked is an example of which of Newton's laws? Newton's second law

Newton's second law states that when an object is acted upon unbalanced force, the object has an acceleration, given by the law

F=ma

So, in this case, the ball is kicked and so an unbalanced force is applied to it, and for this reason the ball has an acceleration (in fact, it starts from rest, but then its velocity increases since it starts moving).

5 0
3 years ago
The two blocks in oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip when the ampli
Setler79 [48]

Answer:

0.72

Explanation:

T = Time period of oscillation = 1.5 s

Angular frequency is given as

w = \frac{2\pi }{T}\\w = \frac{2(3.14) }{1.5}\\w = 4.2 rad/s

A = Amplitude of oscillation = 40 cm = 0.40 m

\mu = Coefficient of static friction = ?

a = acceleration of the block

m = mass of the block

Maximum acceleration of the block is given as

a = Aw^{2}

frictional force is given as

f = \mu mg

As per newton's second law

f = ma \\\\\mu mg = ma \\\mu g = a\\\mu g = Aw^{2}\\\mu (9.8) = (0.40)(4.2)^{2}\\\mu = 0.72

8 0
2 years ago
Why might it be necessary to ignore some of the data points just before and just after the collision?
krek1111 [17]
We have no idea. We need to examine the experimental set-up. You've given us no information, except that there may have been some sort of collision.
7 0
3 years ago
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