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masha68 [24]
2 years ago
5

A 600 kg rocket sled can be accelerated at a constant rate from rest to 1400 km/h in 2.1 s. What is the magnitude of the require

d net force
Physics
1 answer:
kodGreya [7K]2 years ago
8 0

Answer:

7.2 × 10^5 N

Explanation:

The first step is to convert 1400 km/hr to m/s

= 1,400×1000/3600

= 1,400,000/3600

= 388.88 m/s

The acceleration can be calculated as follows

a= v-u/t

= 388.88/2.1

= 185.18

Therefore the required net force can be calculated as follow

= 388.88 × 185.18

= 7.2 × 10^5 N

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A load of 500 N is placed 8 N from the pivot what is the turning moment of the load
Nataliya [291]
The answer is 4000N...
6 0
3 years ago
A sphere is charged with electrons to -6 x 10^-6C. How many electrons make up this charge? The elemental charge is 1.6 x 10^-19
just olya [345]

Answer:

3.75*10^{-13}  electrons

Explanation:

The total charge Q is the sum of the charge of the N electrons contained in the sphere:

Q=N*q_{e}

q_{e}=-1.6*10^{-19}C    charge of a electron

We solve to find N:

N=\frac{Q}{q_{e}}=\frac{-6 x 10^{-6}}{-1.6 x 10^{-19}}=3.75*10^{-13}

7 0
3 years ago
An automobile with tires that have a radius of 0.260 m travels 76000 km before wearing them out. How many revolutions do the tir
Tomtit [17]

Answer:

Explanation:

circumference of the tyre = 2πr = 2 x 3.14 x 0.26 = 1.6328m

76000km = 76000000m

no of revolutions required

= 76000000/1.6328 = 46546 revolutions.

3 0
3 years ago
You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?
Natali5045456 [20]
Equation of motion:

y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}

Since initial velocity is zero, the second term goes away:

y_{f}=y_{o}+0+ \frac{1}{2} at^{2}

y_{f}=y_{o}+\frac{1}{2} at^{2}

y_{f}-y_{o}= \frac{1}{2} at^{2}

y_{f}-y_{o}=5.4m

5.4m= \frac{1}{2} at^{2}

\frac{2(5.4m) }{a} = t^{2}

a = g = 9.81  \frac{m}{ s^{2}}

\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}

1.1 s^{2} = t^{2}\sqrt{1.1 s^{2}} =  \sqrt{t^{2}}

<u><em>t = 1.05s</em></u>
4 0
3 years ago
Read 2 more answers
~~~~~NEED HELP ASAP~~~~~
Romashka-Z-Leto [24]

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

8 0
2 years ago
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