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Artemon [7]
3 years ago
7

A 590-kg rocket is at rest on the launch pad. what upward thrust force is needed to accelerate the rocket uniformly to an upward

speed of 28 m/s in 3.3 s?
Physics
2 answers:
fgiga [73]3 years ago
5 0

<u>Answer:</u> The upward thrust force needed is 5003.2 N

<u>Explanation:</u>

Acceleration is defined as the rate of change of velocity with respect to time.

Mathematically,

a=\frac{v-u}{t}

where,

v = final velocity = 28 m/s

u = initial velocity = 0 m/s

t = time taken = 3.3 s

Putting values in above equation, we get:

a=\frac{28-0}{3.3}=8.48m/s^2

Force is defined as the push or pull on an object with some mass that causes change in its velocity.

It is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

F=m\times a

where,

F = force exerted on the rocket = ?

m = mass of the rocket = 590 kg

a = acceleration of the rocket = 8.48m/s^2

Putting values in above equation, we get:

F=590kg\times 8.48m/s^2\\\\F=5003.2N

Hence, the upward thrust force needed is 5003.2 N

BARSIC [14]3 years ago
3 0
V(final)=V₀+a(t), V(final)=28, V₀=0, t=3.3, solve for a. a=28/3.3=8.5m/s^2. F=ma, F=?, m=590kg, a=8.5m/s^2, F=5015N. 
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A 2000 kg truck is traveling at 5 m/s and collides with a 1000 kg car that is not moving. After the collision, the 2000 truck st
sp2606 [1]

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A) 10 m/s

Explanation:

We know that according to conservation of momentum,

m1v1 + m2v2 = m1u1 + m2u2  ..............(equation 1)

where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.

From the given data

If truck and car are two bodies

truck :       m1 = 2000 Kg           v1 = 5 m/s                u1 = 0

car    :        m2 = 1000 kg           v2 = 0                      u2 = ?

final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.

substituting the values in equation 1, we get

(2000 x 5) + 0 = 0 + (1000 x u2)

u2 = \frac{2000}{1000} x 5

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Two radio antennas are 120 m apart on a north-south line, and they radiate in phase at a frequency of 3.4 MHz. All radio measure
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Answer:

the smallest angle from the antennas is <em>47.3°</em>

Explanation:

We first need to write the expression for the relation between the wavelength (λ) and the frequency (f) of the wave, and then solve for the wavelength.

Therefore, the relation is:

λ = c /f

where

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λ = (3 × 10⁸ m/s) / (3.4 MHz)

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Therefore, the smallest angle measured (from the north of east) from the antennas for the constructive interference of the two-radio wave can be calculated as

θ = sin⁻¹(λ / d)

where

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Thus,

θ = sin⁻¹(88.235 / 120)

<em>θ = 47.3 °</em>

<em></em>

Therefore, the smallest angle from the antennas, measured north of east, at which constructive interference of two radio waves occurs is <em>47.3 °</em>.

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