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Crazy boy [7]
4 years ago
9

A constant force is exerted for a short time interval on a cart that is initially at rest on an air track. This force gives the

cart a certain final speed. Suppose we repeat the experiment but, instead of starting from rest, the cart is already moving with constant speed in the direction of the force at the moment we begin to apply the force.
After we exert the same constant force for the same short time interval, the increase in the cart's speed:

A. is equal to two times its initial speed.
B. is equal to the square of its initial speed.
C. is equal to four times its initial speed.
D. is the same as when it started from rest.
E. cannot be determined from the information provided.
Physics
1 answer:
Ilya [14]4 years ago
5 0

Answer:

d) is the same as when it started from rest

Explanation:

using equation of motion

v = u + at

second law of momentum defines

F = ma

a = F /m

the equation becomes

v = u + (F/m)t

from hear

since v is directly proportional to the force and the force remain the same, the increase in the cart speed will also remain the same.

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A point charge q1 = 3.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 10 m.
UkoKoshka [18]

Answer:

a) 1.6 mN  b) -1.6 mN  c) -1.6 mN  d) 1.6 mN

Explanation:

The electrostatic force between 2 point charges, obeys the Coulomb's Law, that can be expressed as follows:

F₁₂ = k*q₁*q₂/(r₁₂)² (in magnitude)

The direction of the force, is along the  line that joins the  charges (along the x axis) and as q₁ and q₂ are of the same sign, aims away from both charges.

a) So, for the force on q₂, we have:

F₁₂ = 9*18*10⁻⁵ N = 1.6 mN (positive as it is aiming in the positive x direction)

b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:

F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)

c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:

F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)

F₂₁ =  1.6 mN (going in the positive x direction, towards q₂)

6 0
4 years ago
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Floodplains are most often found for rivers that exist on
Kazeer [188]
I am pretty sure that floodplains are most often found for rivers that exist on <span>
hilly areas at the base of mountains. In order to give yoy ans example which will make sure that this answer is quite a suitable one, nice example of f</span><span>loodplains</span>
is The Virgin River<span> at the upper end of Zion Canyon. It will definitely help you! Regards.</span><span>

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3 0
3 years ago
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How does the force of gravity and the force of earth contribute to africa's poverty?
ivanzaharov [21]

Answer:

The force of gravity is not the same as being on the earth. when your on the earth there no gravitational pull its all up to the air

Explanation:

No explanation

8 0
3 years ago
The energy of a photon was found to be 3.38 x 10-19 J. Planck's constant is 6.63 x 10-34 J. S. Which color of light corresponds
GalinKa [24]

Answer:

The correct option is (c).

Explanation:

Given that,

The energy of a photon is, E=3.38\times 10^{-19}\ J

We need to tell the color of this light. We know that, the energy of a photon is given by :

E=\dfrac{hc}{\lambda}

Where

c is the speed of light

\lambda=\dfrac{hc}{E}\\\\\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3.38 \times 10^{-19}}\\\\\lambda=5.88\times 10^{-7}\ m\\\\\lambda=588\ nm

The wavelength of yellow light is approx 580 nm. Hence, we can say that this photon corresponds to yellow light.

6 1
3 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0
2 years ago
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