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3 years ago

Freefalling object starts from rest after falling for six seconds what will be its velocity

Physics

1 answer:

const2013 [10]3 years ago

7 0

Answer:

60m/s

Explanation:

since the object is falling under the influence of gravity, the formula is

v = u +gt

before the fall, the object is at rest then

u = 0m/s

v = 0 + 10×6 because acceleration due to gravity is g which is given as 10m/s2

therefore, v = 60m/s

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3 years ago

Read 2 more answers

If the normal force exerted by the track on the car when it is at the top of the track (point B) is 6.00 N , what is the normal

Eddi Din [679]

Answer:

So, the force, F is the agent which provides the basic property of motion or rest to the body.While, the car is more obviously to have a mass, m and that the angle withe road or surface is also given which is normal to the road(i.e angle =90 degree). Then we say that lets say that the car is moving with the constant velocity of 20 m/sec and its kept unchanged by the car. So, we have the mass, m as 1 kg for the car and the value of the gravity we have the g=9.8 m/sec.

Now,

We have F=ma,

and a=v/t,

so we can have another equation for it as,

Now, providing the required data to, it; ∴t =2 sec,

F=(1)×(20/2),

<u>F=10 N.</u>

So, the car would be acting the force,F of about 10 N while the car is present on the lower region of the track.

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3 years ago

Which interaction between x-ray photons and matter results in total absorption of the incident photon?

Leno4ka [110]

Answer:

Photoelectric effect

Explanation:

In the photoelectric effect, when an x-ray strikes on a metal surface, the energy is completely absorbed by the metal. If the energy would be equal to or more than work function of metal, electron ejects out. The kinetic energy of the electron which is ejected depends on the energy of the incident radiation and work function of the metal.

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4 years ago

Is shooting a a basketball force

lyudmila [28]

Answer:

yes

Explanation:

because when you shot the ball your applying force to it

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4 years ago

Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car

lapo4ka [179]

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

$A_a(t) = 5m/s^2$

To get the velocity, we integrate over time:

$V_a(t) = (5m/s^2)*t + V_0$

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

$V_a(t) = (5m/s^2)*t$

To get the position equation we integrate again over time:

$P_a(t) = 0.5*(5m/s^2)*t^2 + P_0$

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

$P_a(t) = 0.5*(5m/s^2)*t^2$

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

$V_b(t) =20m/s$

To get the position equation, we can integrate:

$P_b(t) = (20m/s)*t + P_0$

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

$P_b(t) = (20m/s)*t + 100m$

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

$P_a(t) = P_b(t)$

We can solve this for t.

$0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0$

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

$t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}$

We only care for the positive solution, which is:

$t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s$

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

$P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m$

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4) What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

$V_a(t) = (5m/s^2)*11.48s = 57.4 m/s$

7 0

3 years ago