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aalyn [17]
1 year ago
11

A 0.56 kg model rocket produces 53 newtons of upward thrust as it shoots upward off the launch pad. With what acceleration meter

s per second-squared does the rocket launch?
Physics
1 answer:
Yanka [14]1 year ago
6 0

resultant force = thrust – weight

acceleration = resultant force (newtons, N) divided by mass (kilograms, kg).

Acceleration = resultant force divided by mass

53N/0.56

=94.64 approximately 95

= 95m/s^2

This means that, every second, the speed of the rocket increases by 95m/s2

the S.I unit of Acceleration is meter per second square.

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José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo
zaharov [31]

r = radius of the circle of the ride = 3.00 meters

v = linear speed of the person during the ride = 17.0 m/s

m = mass of the person in angular motion in the ride

L = angular momentum of the person in the ride = 3570 kg m²/s

Angular momentum is given as

L = m v r

inserting the values

3570 kg m²/s = m (17 m/s) (3.00 m)

m = 3570 kg m²/s/(51 m²/s)

m = 7 kg

hence the mass comes out to be 7 kg


8 0
2 years ago
A rock is thrown straight up with an initial velocity of 19.6 m/s. What time interval elapses between the rock’s being thrown an
inna [77]

Answer:

It will take 4 sec rock to comes its original point

Explanation:

It is given that the rock comes to its original point

So displacement S = 0 m

Initial velocity u = 19.6 m/sec

Acceleration due to gravity g=9.8m/sec^2

According to second equation of motion h=ut+\frac{1}{2}gt^2

0=19.6\times t+\frac{1}{2}\times 9.8t^2

19.6=4.9t

t = 4 sec

3 0
2 years ago
Read 2 more answers
A motorcycle of mass 100 kilograms travels around a flat, circular track of radius 10 meters with a constant speed of 20 meters
JulijaS [17]

Answer:

100/10 = 10 , 10 × 10 = 100÷20 = 5

I'm pretty sure its wrong

8 0
2 years ago
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In a hyrdo electic power plant the moving water turns the turbine creating ____________ energy.
Marrrta [24]

Answer:

....................

Explanation:

................

3 0
2 years ago
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A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At
GuDViN [60]

Answer:

E/4

Explanation:

The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

Where;

E is the electric field

σ is the surface charge density

ε₀ is the electric constant.

Formula to calculate σ is;

σ = Q/A

Where;

Q is the total charge of the sheet

A is the sheet's area.

We are told the elastic sheet is a square with a side length as d, thus ;

A = d²

So;

σ = Q/d²

Putting Q/d² for σ in the electric field equation to obtain;

E = Q/(2ε₀d²)

Now, we can see that E is inversely proportional to the square of d i.e.

E ∝ 1/d²

The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.

From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;

E_new = E/4

3 0
2 years ago
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