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3 years ago

Calculate the acceleration of a car that can go from rest to 100 km/h in 10 seconds

Physics

1 answer:

Basile [38]3 years ago

8 0

Answer: $2.77 m/s^{2}$

Explanation:

Acceleration $a$ is defined as the variation of velocity $\Delta V$ with time $\Delta t$ :

$a=\frac{\Delta V}{\Delta t}$

Where:

$\Delta V=100km/h-0km/h=100km/h=27.77m/s$

$\Delta t=10 s$

$a=\frac{27.77m/s}{10 s}$

Finally:

$a=2.77m/s^{2}$

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A rocket has landed on Planet X, which has half the radius of Earth. An astronaut onboard the rocket weighs twice as much on Pla

Burka [1]

Explanation:

Let acceleration due to Gravity for a planet is given by:

$g_X=GM/R^2$

Here, $g_X = 2g$

Escape velocity is given by:

$v =\sqrt{ \frac{2GM} {R}} = \sqrt{2aR}$

Here, $R=R_earth/2$

and g_X = 2g

Therefore, $v=\sqrt(2(2g)(R/2))=v_0$

7 0

2 years ago

Describe a technology used in space exploration.

Alex777 [14]

Answer:

High speed optical communication technology

To be able to communicate from the space to the earth and from earth to space is one of the most essential features required during space exploration.

Explanation:

Space exploration involves going into the space, beyond the earth's atmosphere. Landing on other planets and studying their details, going into deeper space beyond the planets to discover new cosmic events or structures is all a part of space exploration.

The key to analyse the studies and observations is being able to communicate the data collected, photos taken etc to the launch centers or space centers on earth. The space centers on earth should also be able to communicate with the persons or the satellites in space.

This is made possible using the optical communication technology which involves the use of optical fibers, lasers etc, since high speeds are more efficient during communication

3 0

3 years ago

a baseball player is dashing toward home plate with a speed of 200 feet per hour. When he decides to hit the dirt. he slides for

Serggg [28]

To solve this problem we will use the kinematic formula for the final velocity.

$V_f_x = V_0_x + a_xt$

The final speed is 0 at the moment the player stops.

The time until it stops is 1.3 s

The initial speed is 200 feet / s Note (check the speed units in the problem statement, 200ft / s is very much and 200ft / h is very small)

Then, we clear the formula.

$a_x = \frac{(Vfx-V0x)}{t}\\ a_x = \frac{(0-200)}{1.3}\\ a_x = -153.5 ft / s ^ 2$

Because the player is slowing down, the acceleration goes in the opposite direction to the player's movement, and that is why it is negative.

To answer part b) we use the following formula.

$Vf ^ 2 = Vo ^ 2 + 2a * (x_2 - x_1)\\\\ (x_2 - x_1)= \frac{(V_f ^ 2-V_0 ^ 2)}{2a}\\\\ (x_2 - x_1)= \frac{(0-200 ^ 2)}{- 2 * 153.5}\\\\ (x_2 - x_1)= 130.29 feet$