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Lady bird [3.3K]
2 years ago
11

A 958-N rocket is coming in for a vertical landing. It starts with a downward speed of 25 m/s and must reduce its speed to 0m/s

in 8.0 seconds for the final landing. During this landing maneuver, what must be the thrust due to the rockets engines?
Physics
1 answer:
Art [367]2 years ago
3 0

Answer:

T=1263.16N

Explanation:

A 958-N rocket is coming in for a vertical landing. It starts with a downward speed of 25 m/s and must reduce its speed to 0m/s in 8.0 seconds for the final landing. During this landing maneuver, what must be the thrust due to the rockets engines?

force is that which tends to cane a body state of rest or uniform motion in a straight line

it is also the pull or pus on an object.

the forces involved here to get the thrust on an object, are the weight and the force of gravity

W=mg

958=m*9.81

m=958/9.81

m=97.655kg

T=m(g+a)................................1

to get a

v=u+at

a=dv/dt

a=25/8

a=3.125m/s^2

acceleration is change in velocity per time

juxtaposing into the equation 1

T=97.655(9.81+3.125)

T=1263.16N

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ΔTf = Kf×m
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Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
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7 0
2 years ago
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A carmaker has designed a car that can reach a maximum acceleration of 12 meters/second2. The car’s mass is 1,515 kilograms. Ass
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A visitor to the observation deck of a skyscraper manages to drop a penny over the edge. As the penny falls faster, the force du
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A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved
MariettaO [177]

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4.17 m/s

Explanation:

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u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s

Now we can solve the problem by applying the suvat equation:

v_y^2-u_y^2=2as

where

v_y is the vertical velocity when the pea hits the ceiling

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Instead, the horizontal velocity remains constant during the whole motion, and it is given by

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5 0
2 years ago
The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha
saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

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F = kx

since,

w = Fd

dw = Fdx

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ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

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Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm =  4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0
2 years ago
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