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valentinak56 [21]
4 years ago
11

On a test flight, a rocket with mass 400 kg blasts off from the surface of the earth. The rocket engines apply a constant upward

force F until the rocket reaches a height of 100 m and then they shut off. If the rocket is to reach a maximum height of 400 m above the surface of the earth, what value of F is required? Assume the change in the rocket's mass is negligible.
Physics
1 answer:
Tatiana [17]4 years ago
3 0

Answer:F=11.76 kN

Explanation:

Given

mass of rocket m=400 kg

Force F is applied up to a height of 100 m then it is shut off

final velocity v up to which force is applied is given by

v^2-u^2=2 as

v^2-0=2\times a\times 100

v=\sqrt{200 a}

After this rocket continues to move to a maximum height under the action of gravity

using

v_1^2-v^2=2 gs_2

s_2=400-100=300 m

0-200 a=-2\times 9.8\times 300

a=3 g=3\times 9.8=29.4 m/s^2

thus Force is F=ma

F=400\times 29.4=11,760=11.76 kN

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3 years ago
Solve these problems on charges by finding out Q1 and Q2. Q1Q2= 4x10^6, Q1 + Q2 =8.00×10^-6​
inna [77]

Answer:

Answer:

Q_1 = 7Q

1

=7

Q_2 = 10Q

2

=10

Q_3 = 13.5Q

3

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Step-by-step explanation:

Given

5, 7, 7, 8, 10, 11, 12, 15, 17.

Required

Determine Q1, Q2 and Q3

The number of data is 9

Calculating Q1:

Q1 is calculated as:

Q_1 = \frac{1}{4}(N + 1)Q

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=

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Substitute 9 for N

Q_1 = \frac{1}{4}(9 + 1)Q

1

=

4

1

(9+1)

Q_1 = \frac{1}{4}*10Q

1

=

4

1

∗10

Q_1 = 2.5th\ itemQ

1

=2.5th item

This means that the Q1 is the mean of the 2nd and 3rd data.

So:

Q_1 = \frac{1}{2}(7+7)Q

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Q_1 = \frac{1}{2}*14Q

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1

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Q_1 = 7Q

1

=7

Calculating Q2:

Q2 is calculated as:

Q_2 = \frac{1}{2}(N + 1)Q

2

=

2

1

(N+1)

Substitute 9 for N

Q_2 = \frac{1}{2}(9 + 1)Q

2

=

2

1

(9+1)

Q_2 = \frac{1}{2}*10Q

2

=

2

1

∗10

Q_2 = 5th\ itemQ

2

=5th item

Q_2 = 10Q

2

=10

Calculating Q3:

Q3 is calculated as:

Q_3 = \frac{3}{4}(N + 1)Q

3

=

4

3

(N+1)

Substitute 9 for N

Q_3 = \frac{3}{4}(9 + 1)Q

3

=

4

3

(9+1)

Q_3 = \frac{3}{4}*10Q

3

=

4

3

∗10

Q_3 = 7.5th\ itemQ

3

=7.5th item

This means that the Q3 is the mean of the 7th and 8th data.

So:

Q_3 = \frac{1}{2}(12+15)Q

3

=

2

1

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Q_3 = \frac{1}{2}*27Q

3

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Q_3 = 13.5Q

3

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4 0
3 years ago
Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulle
tangare [24]

Answer:

F=1.159

Explanation:

From the question we are told that:

Mass of pulley M=1kg

Radius r=12cm

Mass of block A M_a=2.1kg

Mass of block B m_b=4.1kg

Spring constant\mu= 358 J/m2

Generally the equation for Torque is mathematically given by

Since \sumF=ma

At mass A

 T_2-f_3=2.1a

At mass B

 4.8-T_1=4.1a

At  Pulley

 R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}

 R(T_1-T_2)=0.55a

Therefore the equation for total force F

At mass A+At mass B+At  Pulley

 (T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a

 (T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a

 -f_3+4.1=6.75a

 -f_3=6.75a+4.8

Since From above equation

M_{eff}=6.7kg

Therefore

T=2\pi \sqrt{{\frac{M_{eff}}{k}}

T=2\pi \sqrt{{\frac{6.75}{\mu}}

T=0.862s

Generally the equation for frequency is mathematically given by

F=\frac{1}{T} \\F=\frac{1}{0.862}

F=1.159

3 0
3 years ago
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STALIN [3.7K]

Answer:

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Explanation:

7

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5 0
3 years ago
A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0 m high cliff. At the instant the ball i
scoray [572]

Answer:

The distance traveled by the woman is 34.1m

Explanation:

Given

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yo = 45m final, positition y = 0m bottom of the cliff

y = yo + ut -1/2gt²

u = 20.0m/s initial speed

g = 9.80m/s²

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Solving quadratically or by using a calculator,

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s = vt

Given the speed of the woman v = 6.00m/s

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s = 6.00×5.69 = 34.14m

Approximately 34.1m to 3 significant figures.

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