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Naya [18.7K]
4 years ago
13

An elevator ascends at a constant speed of 4 m/s, how much time is required for the elevator in order to travel 120 m upwards?

Physics
1 answer:
vlabodo [156]4 years ago
3 0

We simply use the formula,

velocity =\frac{distance}{time}

Given, velocity = 4 m per s and distance = 120 m.

Substituting these values, we get

4\ m/s =\frac{120\ m}{time}\\\\\ time=\frac{120\ m}{4\ m/s} = 30\ s.

Thus, required time for the elevator in order to travel 120 m upwards is 30 s.

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A 98.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.69 r
sertanlavr [38]

Answer:

Explanation:

The problem is based on conservation of angular momentum.

Moment of inertia of the disc = 1/2 m R² , m is mass of the disc and R is its radius.

= 1/2 x 98.1 x 1.51²

= 111.84 kg m²

Moment of inertia of disc + moment of inertia of bananas + monkey

= 1/2 x 98.1 x 1.51² + 9.29 x .45 x 1.51 + 20.3 x 1.51² ( moment of inertia of banana and monkey will be equal to mass x radial distance from axis² )

= 111.84 + 6.31 +46.28

= 164.43 kg m²

Now applying law of conservation of angular momentum

= I₁ ω₁ = I₂ω₂

111.84 x 1.69 = 164.43 x ω₂

ω₂ = 1.15 rad / s

6 0
4 years ago
You are in a fire truck heading directly toward a tunnel in a mountainside at 17m/s. you hear the truck's 90Hz siren as well as
Kryger [21]

Hey there!:

Take the speed of sound to be 343m/s.

Direct frequency perceived by observer:

(343 + 17) / 343) * 90Hz = 94.460Hz

Change in frequency = ( 4.460 - 90 ) = 4.460Hz.  

90 - (4.460 x 2) = 81.08 Hz. indirect frequency heard by observer.  

Therefore :

Beat frequency = (94.460 - 81.08) = 13.38Hz.  

Hope this helps!

4 0
4 years ago
A stone is thrown horizontally from the top of a cliff and eventually hits the ground below. A second stone is dropped from rest
xeze [42]

Answer and Explanation:

As per the question:

When the stone is thrown from the cliff top and hits the ground below eventually:

R = v_{o}\sqrt{\frac{2H}{g}}

where

v_{o} = initial velocity

H = height

g = acceleration due to gravity

R = horizontal Range

Now,

(a) Displacement of the stone is given by the horizontal range:

R = v_{o}\sqrt{\frac{2H}{g}}

where

v_{o} = initial velocity

H = height

g = acceleration due to gravity

R = horizontal Range

(b) Speed just prior to the impact is given by the third equation of motion:

v = \sqrt{v_{o}^{2} + 2gH}

where

v = final velocity

(c) Time of flight is given by the second eqn of motion where the initial velocity is considered to be 0 then:

H = v_{o}T + \frac{1}{2}gT^{2}

H = 0.T + \frac{1}{2}gT^{2}

T = \sqrt{\frac{2H}{g}

3 0
4 years ago
An object will maintain a constant speed until a force acts on it true or false
Scorpion4ik [409]
<span>You do not require a force to keep something moving.  You only require a force to get it moving.  Or to stop it moving.  In your everyday experiences, something you get moving seems to come to a stop after you stop pushing it.  It is because there are forces (friction) that make it stop.  Without those forces, the object would just keep moving. So this would mean the answer would be True.</span>
3 0
3 years ago
you are assigned to do some calculations for a movie stunt that involves a car on a straight road. the road pictured above has a
Vlada [557]

Calculate the time it takes for the car to reach the beginning of the hill given that acceleration as already found in part a is 2.89 m/s²

Answer:

5.88 s

Explanation:

Using kinematic equation, v=u+at where v and u are the final and initial velocities respectively, a is acceleration and t is time.

Considering the first part, acceleration is already found as 2.89 m/s and the final velocity is given as 17 m/s while the initial velocity is zero since it is at rest.

Making t the subject of formula then

t=(v-u)/a

Substituting the given figures then

t=(17-0)/2.89=5.8823529411764s

Rounded off, t=5.88 s

6 0
3 years ago
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