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pshichka [43]
2 years ago
10

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 44.1m/s^2 . The acce

leration period lasts for time 7.00s until the fuel is exhausted. After that, the rocket is in free fall. Find the maximum height y_max reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 rm{m/s^2} .
Physics
1 answer:
ollegr [7]2 years ago
7 0
The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0 
y=44.1*100/2 = 2205m 
hence, the speed will be 
v=0 + a*t = 441m/s 
from that height it will just be subjected to the gravitational acceleration 
0=v_acc^2 -2g*y_free 
y_free = v_acc^2/2g = 9922.5m 
<span>y_max = y_acc+y_free = 441+9922.5 =10363.5m</span>
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zhannawk [14.2K]

Answer:

C. less than 950 N.

Explanation:

Given that

Force in north direction F₁ = 500 N

Force in the northwest F₂ = 450 N

Lets take resultant force R

The angle between force = θ

θ = 45°

The resultant force R

R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}

R=\sqrt{500^2+450^2+2\times 450\times 500\times cos\theta}

R= 877.89 N

Therefore resultant force is less than 950 N.

C. less than 950 N

Note- When these two force will act in the same direction then the resultant force will be 950 N.

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3 years ago
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That they travel in a vacuum.  All other waves require a medium in which they wave.
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3 years ago
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Ahmad was moving to the south with v= 10 km/hr ,and Mohammed was moving with half of Ahmad's speed to the North . Write the vect
mafiozo [28]

Answer:

5 north

Explanation:

7 0
2 years ago
A cannonball is fired on flat ground
algol [13]

hmax = 5740.48 m. The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.

This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.

V₀ = 420m/s and θ₀ = 53.0°

So, when the cannonball is fired it has horizontal and vertical components:

V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s

V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s

When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:

Vy = V₀y - g tₐ = 0

tₐ = V₀y/g

tₐ = (335.43m/s)/(9.8m/s²) = 34.23s

Then, the maximum height is reached in the instant tₐ = 34.23s:

h = V₀y tₐ - 1/2g tₐ²

hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²

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3 0
3 years ago
What is the cost of operating a 100 W light bulb for "4 hrs per day" for 30 days, if the cost of electricity is 6 cents kW•h
Sidana [21]

Answer:

the cost of operating the light bulb is 72 cents.

Explanation:

Given;

cost of electricity, C = 6 cents / kW.h

power of light bulb, P = 100 W

time of light power consumption, t = 4 hours per day for 30 days

total time = 4 hours x 30 = 120 hours

Power consumed by the light bulb is calculated as;

P = 100 x 120 = 12000 w.h = 12 kW.h

Cost of power consumption = 6 cents/kWh x 12 kWh

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Therefore, the cost of operating the light bulb is 72 cents.

3 0
3 years ago
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