All categories

4 years ago

A ball is dropped from a height of 2 m. how long will it take this ball to reach the ground?

Physics

1 answer:

alex41 [277]4 years ago

7 0

Answer:

It depends on the weight or mass or how big the ball is.

You might be interested in

A battery has an electric potential of 1.5V and transfers 10.0 C between the two terminals. How much work was done?

bogdanovich [222]

Answer:

15 Joules

Explanation:

work = charge x potential difference

= 10 x 1.5

= 15

8 0

3 years ago

a soccer player kicks a ball. According to Newton's third law of motion, the player's foot exerts a force on the ball. The ball

Leto [7]

Answer:

The forces are exerted on different objects so they are not balanced forces.

Explanation:

8 0

3 years ago

Which object has the least amount of Kinect energy

Angelina_Jolie [31]

Answer: Could you please add the answer choices.

Explanation:

Thank you :)

6 0

3 years ago

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

4 0

3 years ago

Dragsters can actually reach a top speed of 145.0 m/s in only 4.45 s. (a) Calculate the average acceleration for such a dragster

astraxan [27]

Answer:

a) 32.58 m/s²

b) 161.84 m/s

Explanation:

Initial velocity = u = 0

Final velocity = v = 145 m/s

Time taken = t = 4.45 s

s = Displacement of dragster = 402 m

a = Acceleration

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{145-0}{4.45}\\\Rightarrow a=32.58\ m/s^2$

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 32.58\times 402-0^2}\\\Rightarrow v=161.84\ m/s$

The final velocity is greater than the velocity used to find the average acceleration due to the gear changes. The first gear in a dragster has the most amount of toque which means the acceleration will be maximum. The final gears have less torque which means the acceleration is lower here. The final gears have less acceleration but can spin faster which makes the dragster able to reach higher speeds but slowly.

7 0

3 years ago