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3 years ago

A 1200 kg car accelerates from 13m/s to 17m/s. find the change in momentum of the car?

Physics

1 answer:

lisabon 2012 [21]3 years ago

6 0

Answer:

4800

Explanation:

The change in velocity of the car is 17-13=4m/s. Since the change in momentum is the mass multiplied by the change in velocity, the answer is 4*1200=4800. Hope this helps!

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A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in

pantera1 [17]

Answer:

96046 Ns.

Explanation:

We shall represent velocity in vector form considering east direction as + ve x axis and north as + y direction.

40 km/h in the east

V₁ = 40 i

V₂ = 50j

momentum p₁ = mV₁

= 1500 X 40 i

= 60000 i

Momentum p₂ = mV₂

= 1500 X 50j

= 75000 j

Change in momentum

p₂ - p₁

75000j - 60000i

Magnitude of change

= $\sqrt{(750000)^2 +(60000)^2$

= 96046 Ns.

5 0

3 years ago

A certain wave motion has a frequency of 10 cycles / s and a wavelength of 3 m. so its propagation speed is

kodGreya [7K]

3.5m is ur answer ask for more questions anytime

5 0

3 years ago

Rudiy27

Okay so yeah u have to minus then subtract then decide it it’s a method i was taught to do

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3 years ago

How is Coulomb’s law similar to newton’s law of gravitational force? How is it different

natulia [17]

The similarities and the differences between gravitational and electric force are listed below

Explanation:

- The magnitude of the gravitational force between two objects is given by Newton's law of gravitation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

- Coloumb's law gives instead the strength of the electrostatic force between two charged objects, which is

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

By comparing the two equations, we find the following similarities:

Both the forces are inversely proportional to the square of the distance between the two objects, $F\propto \frac{1}{r^2}$
Both the forces are proportional to the product between the "main quantity" of each force, which is the mass for the gravitational force ( $F\propto m_1 m_2$ ) and the charge for the electric force ( $F\propto q_1 q_2$

Instead, we have the following differences:

The gravitational force is always attractive, since the sign of $m$ is always positive, while the electric force can be either attractive or repulsive, since the sign of $q$ can be either positive or negative
The value of the gravitational costant G is much smaller than the value of the Coulomb's constant, so the gravitational force is much weaker than the electric force