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melisa1 [442]
3 years ago
13

Write a brief essay describing how Newton’s Laws explain how a rocket in space can move objects

Physics
2 answers:
statuscvo [17]3 years ago
5 0
The First Law describes how an object acts when no force is acting upon it. So, rockets stay still until a force is applied to move them. Likewise, once they're in motion, they won't stop until a force is applied. Newton's Second Law tells us that the more mass an object has, the more force is needed to move it. A larger rocket will need stronger forces (eg. more fuel) to make it accelerate. The space shuttles required seven pounds of fuel for every pound of payload they carry. Newton's Third Law states that "every action has an equal and opposite reaction". In a rocket, burning fuel creates a push on the front of the rocket pushing it forward.
BaLLatris [955]3 years ago
3 0

Answer:

The First Law describes how an object acts when no force is acting upon it. So, rockets stay still until a force is applied to move them. Likewise, once they're in motion, they won't stop until a force is applied. Newton's Second Law tells us that the more mass an object has, the more force is needed to move it. A larger rocket will need stronger forces (eg. more fuel) to make it accelerate. The space shuttles required seven pounds of fuel for every pound of payload they carry. Newton's Third Law states that "every action has an equal and opposite reaction". In a rocket, burning fuel creates a push on the front of the rocket pushing it forward.

Explanation:

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IgorLugansk [536]

Answer:

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8 0
3 years ago
How much force would be needed to cause a 4.6kg object to accelerate at 9.2m/s/s? *
poizon [28]

Answer:

<h2>42.32 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 4.6 × 9.2

We have the final answer as

<h3>42.32 N</h3>

Hope this helps you

5 0
3 years ago
Two people are standing on a 1.75-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of a
Fiesta28 [93]

Answer:

0.05312 m

Explanation:

Given:

Length of platform L = 1.75 m

mass of ball m_b = 5.76 kg

mass of (people + platform) m_p = 184 kg

Initial Velocity of ball V_i,b = 0

Initial Velocity of ball (people + platform) V_i,p = 0

Find:

How far does the platform recoils to rest

Solution:

Using the conservation of momentum on ust before and after the ball was thrown P_i = P_f :

Where, P_i = 0 (initially at rest)

P_f = m_p*V_f,p + m_b * V_f,b

0 = m_p*V_f,p + m_b * V_f,b

V_f,p = - (m_b /m_p) * V_f,b

V_f,p = - (5.76 / 184)*V_f,b

V_f,p = - 0.0313*V_f,b   ....1

The time the ball is in air:

t = L / (V_f,b - V_f,p)   ...2

The distance that the platform moves d:

d = V_f,p *t  ....3

Substitute 2 into 3

d = V_f,p*L /(V_f,b - V_f,p)   .... 4

Solve 1 and 4 simultaneously :

d = - m_b*L / (m_b + m_p)

d = - 5.76*1.75 / (5.76 + 184)

d = -0.05312 m

The platform moves 0.05312 m to the opposite to which the ball is thrown.

5 0
3 years ago
A 1.5 kg ball is dropped from a height of 2.Gm. Assuming energy is
Vikki [24]

Answer:

Plug in the given values and solve for the final velocity. Remember, when the ball is on the ground it has a height of zero.

Explanation:

6 0
2 years ago
The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t
Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

PE=\frac{GMm}{r}

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

m = \rho V

m = (1030Kg/m^3)(7*10^8m^3)

m = 7.210*10^{11}Kg

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m

PART A) Potential energy when the ocean is at its furthest point to the moon,

PE_1 = \frac{GMm}{r_1}

PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}

PE_1 = 9.05*10^{15}J

PART B) Potential energy when the ocean is at its closest point to the moon

PE_2 = \frac{GMm}{r_2}

PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}

PE_2 = 9.361*10^{15}J

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

\Delta KE = \Delta PE

\frac{1}{2}mv^2 = PE_2-PE_1

v=\sqrt{2(PE_2-PE_1)/m}

v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}

v = 29.4m/s

8 0
4 years ago
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