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allsm [11]
3 years ago
6

A rocket, weighing 4.36 x 10^4 N, has an engine that provides an upward force of 8.90 x 10^5 N. It reaches a maximum speed of 86

0 m/s. What is the NET force on the rocket
Physics
1 answer:
Pavel [41]3 years ago
8 0

The net force on the rocket is  846400 N.

Answer:

Explanation:

It is known that weight is the influence of gravitational force acting on any mass of the object. So in the present case, the weighing force given is equal to the gravitational force acting on the rocket. Thus, the gravitational force will be acting towards downward direction. But an upward force is required by the rocket for thrusting purpose and that force is given as upward force. So the net force acting on the rocket is the vector addition of all the forces acting on the rocket. As in this case, only upward and downward force is acting on the rocket. The vector addition will be equal to subtraction of downward acting gravitational force from upward force or force provided by engine.

Net force = Engine force - Gravitational force = 890000-43600=846400 N

So the net force acting on the rocket is 846400 N.

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TheE-field strength is 50,000 N/C inside a parallel plate capacitor with 2.0 mmspacing. A proton is released from rest at the po
Ghella [55]

Answer:

The speed is 138852.4 \frac{m}{s}

Explanation:

Electric field magnitude (E) and electric force magnitude (F) are related by the equation

F=Eq

with q the charge of the proton (1.61\times10^{-19} C). Because between parallel plates electric field is almost constant, electric force is constant too:

F=(50000)(1.61\times10^{-19})=8.05\times10^{-15}N

because electric force is constant, then by Newton's second law acceleration (a) is constant too, it is:

a=\frac{F}{m}

with m the mass of the proton (1.67\times10^{-27} kg):

a=\frac{8.05\times10^{-15}}{1.67\times10^{-27}}= 4.82\times10^{12}\frac{m}{s^{2}}

Now with this constant acceleration we can use the kinematic equation

v^{2}=v_{0}^{2}+2ad

with v the final speed, v_i the initial velocity that is zero (because proton starts at rest) and d is the distance between the plates, so:

v=\sqrt{2ad}=\sqrt{2(4.82\times10^{12}\frac{m}{s^{2}})(2.0\times10^{-3}m)}

v=138852.4 \frac{m}{s}

5 0
3 years ago
A 75.5 kg diver drops from a diving board 10.0 m above the waters surface. Find the divers speed just before he strikes the wate
jenyasd209 [6]

Answer:

<em>v=14 m/s</em>

Explanation:

<u>Mechanical Energy </u>

The kinetic energy of a body (K) is the capacity of doing work due to its speed. It can be expressed as

\displaystyle K=\frac{mv^2}{2}

The potential energy (U) is the capacity of doing work due to its height respect to a certain reference level.

U=mgh

The  mechanical energy is the sum of both

\displaystyle E_m=\frac{mv^2}{2}+mgh

The principle of conservation of mechanical energy states it must remain the same if no external force is acting on it. The diver drops from the diving board, which means its initial speed is zero (and so its initial kinetic energy). Thus, the mechanical energy at the jumping time is

\displaystyle E_m=mgh=(75)(10)(9.8)=7350\ J

When the diver is about to get into the water, his height reaches zero and the speed is at maximum. All the potential energy became kinetic energy, so

\displaystyle \frac{mv^2}{2}=7350\ J

Rearranging

\displaystyle v^2=\frac{2(7350)}{75}=196

v=\sqrt{196}=14\ m/s

The final speed of the diver is

\boxed{v=14\ m/s}

8 0
3 years ago
A bird is standing on an electric transmission line carrying 3000 A of current. A wire like this has about 3.0 x 10-5 22 of resi
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Answer:

13.5 x 10^-9 A

Explanation:

Yes

6 0
3 years ago
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