What is the force on a 1,000-kilogram elevator that is falling under the acceleration of gravity?

A certain light truck can go around a flat curve having a radius of 150 m with a maximum spee dof 32.0 m/s. With what maximum sp

Dvinal [7]

Answer

Maximum speed at 75 m radius will be 22.625 m /sec

Explanation:

We have given radius of the curve r = 150 m

Maximum speed $v_{max}=32m/sec$

Coefficient of friction $\mu =\frac{v_{max}^2}{rg}=\frac{32^2}{150\times 9.8}=0.6965$

Now new radius r = 75 m

So maximum speed at new radius $v_{max}=\sqrt{\mu rg}=\sqrt{0.6965\times 75\times 9.8}=22.625m/sec$

7 0

3 years ago

A ball is dropped from an aircraft flying at an altitude of 8,848 meters assuming gravity is 9.8m/s what is the total amount of

Dafna11 [192]

In this question, you're determining the time (t) taken for an object to fall from a distance (d).

The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)

d = 8,848m and g = 9.8m/s

Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)

The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)

Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)

Now for the last step, find the square root of the remaining number:
t = 42.5s

So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.

I hope this helps :)

7 0

3 years ago

A ball is thrown horizontally from a 19 m -high building with a speed of 2.0 m/s . How far from the base of the building does th

bazaltina [42]

Using the formula: ΔY = V₀y * t + (1/2) * ay * t²

Solve for time and get: 1.968s

Then use: v = d/t in the x-direction and get: d = 3.936

3 0

3 years ago

Do Earth's continents gain or lose water, considering evaporation and precipitation together? How much ? Plz help :)) Thx

ratelena [41]

Answer:

<em>The amount of water in circulation is the same</em>

Explanation:

The water cycle is the pathway in which water moves from the atmosphere to the earth and back to atmosphere ones again.

During the sunny period, the temperature of the earth is heated up, water evaporates to the atmosphere as water vapour. The water vapour gathers as cloud and comes back to the earth as rain, which enters our rivers and stream. Thereby maintaining a continuous cycle.

With the water cycle, the movement of water via precipitation and evaporation is the same on earth.

5 0

3 years ago

Conservation of Momentum Practice

Yakvenalex [24]

Answer:

Before: 0 m/s

After: -4 m/s

Explanation:

Before: Since you and your beau started at rest, your beau initial velocity is 0 m/s.

After: Since we have to conserve momentum,

momentum before push = momentum after push.

The momentum before push = 0 (since you and your beau are at rest)

momentum after push = m₁v₁ + m₂v₂ were m₁ = your mass = 60 kg, v₁ = your velocity after push = 3 m/s, m₂ = beau's mass = 45 kg and v₂ = beau's velocity.

So, m₁v₁ + m₂v₂ = 0

m₁v₁ = -m₂v₂

v₂ = -m₁v₁/m₂ = -60 kg × 3 m/s ÷ 45 kg = -4 m/s

So beau moves with a velocity of 4 m/s in the opposite direction

8 0

3 years ago