Calculate the average acceleration of a car that changes speed from 0 m/s to 20 m/s in 5 s.

Physics Homework

katrin2010 [14]

Explanation:

a. Average speed = distance / time

= 100 m / 70 s

= 1.43 m/s

b. Average displacement = displacement / time

= 0 m / 70 s

= 0 m/s

Distance is the length of the path traveled. Displacement is the difference between the final position and initial position.

4 0

4 years ago

A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

Firstly we find the vertical component of the initial velocity:

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

Now we find the time taken to ascend to this height:

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

Time taken to descent the total height:

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

Now the total time taken by the ball to hit the ground:

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

3 0

3 years ago

A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the

spayn [35]

Answer:

h = 16.9 m

Explanation:

When a ball is thrown upward, its velocity gradually decreases, until it stops for a moment, when it reaches the maximum height, while its height increases. Thus, the law conservation of energy states in this case, that:

Kinetic Energy Lost by Ball = Potential Energy Gained by Ball

(0.5)m(Vf² - Vi²) = mgh

h = (0.5)(Vf² - Vi²)/g

where,

Vf = Final Speed of Ball = 0 m/s (Since, ball stops for a moment at highest point)

Vi = Initial Speed of Ball = 18.2 m/s

g = acceleration due to gravity = - 9.8 m/s² ( negative for upward motion)

h = maximum height the ball can reach = ?

Therefore, using values in the equation, we get:

h = (0.5)[(0 m/s)² - (18.2 m/s)²]/(-9.8 m/s²)

h = 16.9 m

4 0

3 years ago

If anyone can answer it'd be very appreciated, i need to pass this class

alexandr1967 [171]

Answer: 0.4

Explanation:

8 0

3 years ago

Read 2 more answers

A satellite is in circular orbit around the earth. The orbit the satellite is at a height of 420 km above the earth's surface. F

astraxan [27]

Answer:

7661.06 m/s

Explanation:

R = Radius of Earth = $6.38\times 10^6\ m$

h = Distance from the Earth = 420000 m

G = Gravitational constant = $6.674\times 10^{-11} N m^2/kg^2$

M = Mass of Earth = $5.98\times 10^{24}\ kg$

$V=\sqrt{g{\frac{R^2}{R + h}}}\\\Rightarrow V=\sqrt{\frac{GM}{R^2}{\frac{R^2}{R + h}}}\\\Rightarrow V=\sqrt{\frac{6.674\times 10^{-11}\times 5.98\times 10^{24}}{(6.38\times 10^6)^2}{\frac{(6.38\times 10^6)^2}{6.38\times 10^6 + 420000}}}\\\Rightarrow V=7661.06\ m/s$

The orbital speed of the satellite is 7661.06 m/s

6 0

4 years ago