All categories

2 years ago

What are the effects of forces to the motion of an object?

1 answer:

3 0

Forces affect how objects move. They may cause motion; they may also slow, stop, or change the direction of motion of an object that is already moving. Since force cause changes in the speed or direction of an object, we can say that forces cause changes in velocity. Remember that acceleration is a change in velocity.

You might be interested in

Which of the following particles are responsible for producing electrical energy?

kherson [118]

The answer is Electrons

4 0

3 years ago

b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 2.50 m/s

fomenos

Answer:

The horizontal distance is 0.64 m.

Explanation:

Initial velocity, u =2.5m/s

The maximum horizontal distance is

$R = \frac{u^2}{2g}\\\\R = \frac{2.5\times 2.5}{9.8}\\\\R = 0.64 m$

3 0

2 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

7nadin3 [17]

Answer:

(a) r = 1.062·R $_E$ = $\frac{531}{500} R_E$

(b) r = $\frac{33}{25} R_E$

Explanation:

Here we have escape velocity v $_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v $_e$ we have

v = $0.241\times \sqrt{\frac{2GM}{R_E} }$ so that;

v² = $0.058081\times {\frac{2GM}{R_E} }$

v² = ${\frac{0.116162\times GM}{R_E} }$

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

r = 1.062·R $_E$

(b) Here we have

$K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}$

Therefore we put $\frac{0.241GM}{R_E}$ in the maximum height equation to get

$\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}$

From which we get

r = 1.32·R $_E$

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0

2 years ago

Read 2 more answers

) Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10% of ene

Angelina_Jolie [31]

Answer:

8100W

Explanation:

Let g = 10m/s2

As water is falling from 60m high, its potential energy from 60m high would convert to power. So the rate of change in potential energy is

$P = \dot{E} = \dot{m}gh = 15*10*60 = 9000 J/s$ or 9000W

Since 10% of this is lost to friction, we take the remaining 90 %

P = 9000*90% = 8100 W

3 0

2 years ago

A man releases a stone ar the top edge of a tower during the last second of its travel the stone falls through a distance of (9/

Musya8 [376]

Equations of motion (EoM) use EoM v2=u2+2ax to establish velocities at positions shown in blue in drawing from EoM v=u+at for final 1 second of flight time, we can say v=u+g(1) 2gH−−−−√=2g1625H−−−−−−√+g then, solve for H [in terms of g]

4 0

2 years ago

Read 2 more answers

What are the effects of forces to the motion of an object?​

What are the effects of forces to the motion of an object?