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AleksandrR [38]

3 years ago

13

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor sto

ps. The rocket altitude 20 s after launch is 6600 m. You can ignore any effects of air resistance.
Required:
a. What was the rocket's acceleration during the first 16s?
b. What is the rocket's speed as it passes through acloud 5100 m above the ground?

1 answer:

Fed [463]3 years ago

8 0

Answer:

a) a = 34.375 m / s², b) v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

v_f = $\frac{x-x_1}{t}$

we substitute the values

v_f = $\frac{ 6600 -x_1}{4}$

The initial part of the movement is carried out with acceleration

v_f = v₀ + a t

x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

x₁ = ½ a t²

v_f = a t

we substitute the values

x₁ = 1/2 a 16²

x₁ = 128 a

v_f = 16 a

let's write our system of equations

v_f = $\frac{6600 - x_1}{4}$

x₁ = 128 a

v_f = 16 a

we substitute in the first equation

16 a = $\frac{6600 -128 a}{4}$

16 4 a = 6600 - 128 a

a (64 + 128) = 6600

a = 6600/192

a = 34.375 m / s²

b) let's find the time to reach this height

x = ½ to t²

t² = 2y / a

t² = 2 5100 / 34.375

t² = 296.72

t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

v_f = 16 a

v_f = 16 34.375

v_f = 550 m / s

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3 years ago

A parcel has a mass of 500 g .calculate its weight (assume the gravitational field strength is 10 N/Kg

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Answer:

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Explanation

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3 years ago

a hippopotamus produces a pressure of 250000 pa when it is standing on all four feet if the weight of the hippo is 40000 N what

mamaluj [8]

0.04m²

Explanation:

Given parameters:

Pressure = 250000Pa

Weight = 40000N

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Solution:

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To find the area;

Area = $\frac{force }{pressure}$

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8 0

3 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

4 years ago

A 1400 kg car is traveling east on the highway at 31 m/s and collides into the rear of a slower moving pickup truck of 2400 kg,

zloy xaker [14]

Answer: 31 m/s due east

Explanation: this question can be solved using the law of conservation of linear momentum.

This law states that in a closed or isolated system, during collision, the vector sum of momentum before collision equals the vector sum of momentum after collision.

Momentum = mass × velocity

From our question, our parameters before collision are given below as

Mass of car = mc = 1400kg

Speed of car =vc = 31 m/s (due east)

Mass of truck = mt = 2400kg

Velocity of truck = vt = 25 m/s ( due east )

After collision

Velocity of car = ?

Velocity of truck = 34 m/s ( due east )

Vector sum of momentum before collision is given as

1400 (31) + 2400 (25) = 43400 + 60000 = 103400 kgm/s

After collision the truck is seen to move faster (v = 34 m/s) which implies that the car also moves due east .

1400 (v) + 2400(25) .... A positive value is between both momenta because they are in the same direction.

After collision, we have that

1400v + 60000

Vector sum of momentum before collision = vector sum of momentum after collision

103400 = 1400v + 60000

103400 - 60000 = 1400v

43400 = 1400v

v = 43400/ 1400

v = 31 m/s due east

4 0

4 years ago