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3 years ago

What is the magnitude of velocity for a 2,000 kg car possessing 3,000 kg(*)m/s of momentum?

Physics

1 answer:

Trava [24]3 years ago

5 0

The magnitude of velocity for this car is equal to 1.5 m/s.

<u>Given the following data:</u>

Momentum of car = 3,000 kgm/s.

Mass of car = 2,000 kg.

To calculate the magnitude of velocity for this car:

<h3>What is momentum?</h3>

In Science, momentum simply means a multiplication of the mass of an object and its velocity.

Mathematically, momentum is giving by the formula;

$Momentum = mass \times velocity$

Making velocity the subject of formula, we have:

$Velocity=\frac{Momentum}{Mass}$

Substituting the given parameters into the formula, we have;

$Velocity=\frac{3000}{2000}$

Velocity = 1.5 m/s.

Read more on momentum here: brainly.com/question/15517471

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3 years ago

Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22

natka813 [3]

Answer:

2420 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 22.9 N

Angle (θ) = 35°

Distance (d) = 129 m

Workdone (Wd) =?

The work done can be obtained by using the following formula:

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Thus, the workdone is 2420 J.

3 0

3 years ago

A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force

Tom [10]

Answer:

$VB - VA = - 33.4$

Explanation:

Generally the workdone in moving the proton is mathematically represented as

$W = KE_f - KE_i$

Where $KE_i \ and \ KE_f \ are\ the\ initial \ and \ final \ kinetic \ energy$

$KE_i = \frac{1}{2} m v_a^2$

Here $v_a$ is the velocity at A with value 50 m/s

$KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2$

$KE_i = 2.09 *10^{-24} \ J$

Also

$KE_f = \frac{1}{2} m v_b^2$

Here $v_a$ is the velocity at A with value $80 km/s = 80000 m/s$

=> $KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2$

=> $KE_f = 5.34 *10^{-18} \ J$

$W = 5.34 *10^{-18} - 2.09 *10^{-24}$

$W = 5.34 *10^{-18} m/s$

Now this workdone is also mathematically represented as

$W = q * V$

$q * V = 5.34 *10^{-18}$

Here $q = 1.60*10^{-19} C$

$V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}$

$V = 33.4 \ V$

Generally proton movement is in the direction of the electric field it means that $VA>VB$

$VB - VA = - 33.4$

8 0

3 years ago

Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S

Alja [10]

Answer:

16294 rad/s

Explanation:

Given that

M(ns) = 2M(s), where

M(s) = 1.99*10^30 kg, so that

M(ns) = 3.98*10^30 kg

Again, R(ns) = 10 km

Using the law of gravitation, the force between the Neutron star and the sun is..

F = G.M(ns).M(s) / R²(ns), where

G = 6.67*10^-11, gravitational constant

Again, centripetal force of the neutron star is given as

F = M(ns).v² / R(ns)

Recall that v = wR(ns), so that

F = M(s).w².R(ns)

For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence

F = F

G.M(ns).M(s) / R²(ns) = M(s).w².R(ns)

Making W subject of formula, we have

w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]

w = √[{G.M(ns)} / {R³(ns)}]

w = √[(6.67*10^-11 * 3.98*10^30) / 10000³]

w = √[2.655*10^20 / 1*10^12]

w = √(2.655*10^8)

w = 16294 rad/s

7 0

3 years ago