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Mumz [18]
3 years ago
7

What is the average acceleration of a rocket that travels 400m in 5s if it starts from rest?​

Physics
1 answer:
Pavel [41]3 years ago
4 0

Answer:

<h2>32m/s^2</h2>

Explanation:

We want to find the acceleration based on the given data

Given

distance s=400m

time t= 5s

u= 0m/s since it started from rest

We apply the following expression

s=ut+1/2at^2

substituting we have

400=0*5+1/2*a(5)^2

400=25a/2

cross multiply

25a=400*2

25a=800

divide both sides by 25 we have

a=800/25

a=32m/s^2

The acceleration is 32m/s^2

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Electromagnetic radiation from a 8.25 mW laser is concentrated on a 1.23 mm2 area. Suppose a 1.12 nC static charge is in the bea
mylen [45]

Answer:

The maximum magnetic force is 2.637 x 10⁻¹² N

Explanation:

Given;

Power, P = 8.25 m W = 8.25 x 10⁻³ W

charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C

speed of the charge, v = 314 m/s

area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²

The intensity of the radiation is calculated as;

I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2

The maximum magnetic field is calculated using the following intensity formula;

I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T

The maximum magnetic force is calculated as;

F₀ = qvB₀

F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)

F₀ = 2.637 x 10⁻¹² N

5 0
3 years ago
A cruise ship travels across a river at 25 meters per minute. If the river is 6200 meters wide, how long
EleoNora [17]

Answer:

248 minutes

Explanation:

6200/25=248

This means there is 248 25s in 6200

which means it will take 248 minutes to travel through the river

Also here's a neat trick:

The units for speed is meters/minute

The units for distance is meters

Dividing distance by speed will cancel out the meters and leave only the speed.

3 0
3 years ago
When heat is added to an object, what is happening to the item at the atomic level? (Check all that apply)
3241004551 [841]

Answer:

a

Explanation:

heat is energy, energy cannot be made or destroyed but transferred

6 0
3 years ago
Read 2 more answers
A car traveling at 60km/h undergoes uniform acceleration at a rate of 2/ms^2 until is velocity reached 120km/h determine the dis
jenyasd209 [6]

Explanation:

Given that,

Initial speed of a car, u = 60 km/h = 16.67 m/s

Acceleration, a = 2m/s²

Final speed, v = 120 km/h = 33.33 m/s

We need to find the distance traveled and the time taken to make the distance.

acceleration = rate of change of velocity

a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{33.33 -16.67 }{2}\\\\t=8.33\ s

let the distance be d.

d=\dfrac{v^2-u^2}{2a}\\\\d=\frac{33.33^{2}-16.67^{2}}{2(2)}\\\\d=208.25\ m

Hence, the distance traveled and the time taken to make the distance is 208.25 m and 8.33 seconds respectively.

3 0
3 years ago
A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is  T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2  m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is  mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0
3 years ago
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