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3 years ago

What is the velocity of a ball dropped from a height of 150 m when it hits the ground? Take the upward direction as positive.

Physics

1 answer:

djyliett [7]3 years ago

7 0

Answer:

The velocity of the ball when its hit the ground will be 54.22 m/sec

Explanation:

We have given height from which ball is dropped h = 150 m

Acceleration due to gravity $g=9.8m/sec^2$

As the ball is dropped so initial velocity will be zero so u = 0 m/sec

According to third equation of motion we know that $v^2=u^2+2gh$

$v^2=0^2+2\times 9.8\times 150$

$v=54.22m/sec$

So the velocity of the ball when its hit the ground will be 54.22 m/sec

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Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago

Plzzzzzzzzzzzzzzzzzz helpppp

satela [25.4K]

It’s c, the toy car changes direction

7 0

2 years ago

Read 2 more answers

What's the Coulomb's law?

Ulleksa [173]

In layman's term: like charges don't attract while opposite charges doelectrostatic forces between point A( which is charged) and point B (which is also charged) are proportional to the charge of point A and point B. there is also something else about this law that I don't quite remember.

___________________________________________________

Here is the formula:

F = k x Q1 x Q2/d^2

What the formula means:

F=force between charges

Q1 and Q2= amount of charge

d=distance between these two charges

k= Coulombs constant (proportionally constant)

________________________________________________

I think that about covers it and hopefully this helped.

4 0

2 years ago

Read 2 more answers

A 1000 kg race car rounds a curve with a radius of 50 m.

Kruka [31]

Answer:

5644556677888777766554433

5 0

2 years ago

A metal rod has a moves with a constant velocity of 40 cm/s along two parallel metal rails through a magnetic field of 0.575 T.

love history [14]

Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

= B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

= 0.002116/R W

= 2.12/R mW

3 0

3 years ago