All categories

2 years ago

You drop a rock down a well that is 11.5 m deep.How long does it take the rock to hit the bottom of the well ?

1 answer:

4 0

The acceleration of gravity depends on the distance from
the center of the Earth. We use 9.8 m/s² for any place that's
on the Earth's surface, or reasonably close. The 11.5 meters
down into the well certainly qualifies.

The formula we want is:

Distance of fall = (1/2) · (acceleration) · (time)²

      11.5 m = (1/2) (9.8 m/s²) (time)²

Divide each side
by 4.9 m/s² : (11.5m) / (4.9 m/s²) = time²

   2.347 sec² = time²

   time = √(2.347 sec²) =   1.53 seconds .

You might be interested in

What is the difference between free fall and weightlessness??

joja [24]

When you are in free fall, the force of gravity is stronger than your velocity perpendicular to where you're falling, and you move at a constant speed downwards.

Under feelings of weightlessness, you are still being pulled by gravity, but your perpendicular velocity and distance from the source can cancel each other out.

3 0

2 years ago

Three identical train cars, coupled together, are rolling east at 1.8 m/s . A fourth car traveling east at 4.5 m/s catches up wi

Vedmedyk [2.9K]

Answer:

$v = 1.98\ m/s$

Explanation:

consider the mass of each train car be m

m₁ = m₂ = m₃ = m

speed of the three identical train

u₁ = u₂ = u₃ = 1.8 m/s

m₄ = m u₄ = 4.5 m/s

m₅ = m u₅ = 0 (initial velocity )

final velocity

v₁ = v₂ = v₃ = v₄ = v₅ = v

using conservation of momentum

m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅

m (1.8 + 1.8 + 1.8 +4.5) = 5 m v

$v = \dfrac{9.9}{5}$

$v = 1.98\ m/s$

5 0

2 years ago

During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa

maw [93]

Answer:

Explanation:

Given

Initial speed $u=60\ mi/hr\approx 88\ ft/s$

distance traveled before coming to rest $d_1=120\ ft$

using equation of motion

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$0-(88)^2=2\times a\times 120---1$

for $u_2=80\ mi/hr\approx 117.33\ ft/s$

using same relation we get

$0-(117.33)^2=2\times a\times (d_2)----2$

divide 1 and 2 we get

$(\frac{88}{117.33})^2=\frac{120}{d_2}$

$d_2=213.32\ ft$

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

8 0

2 years ago

A 2 kg ball of putty moving to the right at 3 m/s has a perfectly inelastic, head-on collision with a 1 kg ball of putty moving

Aneli [31]

Answer:

$V=1.33m/s$ to the right

Explanation:

The balls collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V.

We need to use the conservation of momentum Law, the total momentum is the same before and after the collision.

In the axis X:

$m_{1}*v_{o1}-m_{2}*v_{o2}=(m_{1}+m_{2})V$ (1)

$V=(m_{1}*v_{o1}-m_{2}*v_{o2})/(m_{1}+m_{2})=(2*3-1*2)/(2+1)=1.33m/s$

5 0

2 years ago

Kepler's second law, which states that as a planet moves around its orbit it sweeps out equal areas in equal times, means that K

erastovalidia [21]

Explanation:

not physics tho??? idkkkkkkkkk sorrrryyyyyy

3 0

2 years ago