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ArbitrLikvidat [17]
3 years ago
6

What is the speed of a 400 kg elevator that has 1800J of energy

Physics
1 answer:
sveta [45]3 years ago
8 0

Answer: 1400

Explanation:

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A 1400 kg car is moving at 33.8 m/s when a force is applied the opposite direction of the car's motion. The car slows down to 21
zubka84 [21]

Solve for acceleration:

<em>a</em> = (21.4 m/s - 33.8 m/s) / (4.7 s)

<em>a</em> ≈ -2.6 m/s²

Solve for force:

<em>F</em> = (1400 kg) <em>a</em> ≈ -3700 N

The minus sign tells you the force points in the opposite direction of the car's motion. Its magnitude is always positive, so <em>F</em> = 3700 N.

3 0
3 years ago
A student notices that an inflated balloon gets larger when it is warmed by a lamp. Which best describes the mass of the balloon
nirvana33 [79]
It gets larger because
well let me give you an example
so today in class we looked at a lava lamp with wax inside and there was a lightbulb at the bottom.
we watched as the wax floated up because the molecules inside the wax spreads out and makes the wax less dense.
the wax floats up because (which is related to the balloon getting bigger) the wax is getting less dense and the particles get bigger which ALSO makes the wax less dense.
hope this helps and hope you can relate it to your problem! say thanks if I did help AT ALL! :)
7 0
4 years ago
Read 2 more answers
During which two time intervals does the particle undergo equal displacement?
san4es73 [151]

Answer:

BC and DE

Explanation:

In the given figure, the velocity time graph is shown. We know that the area under v-t curve gives the displacement of the particle.

Area under AB, d_1=\dfrac{1}{2}\times 2\times 10=10\ m

Area under BC, d_2=\dfrac{1}{2}\times 2\times 4=4\ m

Area under CD, d_3=\dfrac{1}{2}\times 2\times 7=7\ m

Area under DE, d_4=\dfrac{1}{2}\times 2\times 4=4\ m

Area under EF, d_5=\dfrac{1}{2}\times 2\times 3=3\ m

So, form above calculations it is clear that, during BC and DE undergo equal displacement. Hence, the correct option is (c) "BC and DE = 4 meters".

4 0
3 years ago
an object has a mass of 50kg, a final height of 20m and an initial height of 8m. what is the amount of work done
Andrei [34K]

amount of work done is 5880 J

Given:

mass of object = 50kg

Final height = 20m

initial height = 8m

To Find:

amount of work done

Solution:

work is done when a force acts upon an object to cause a displacement. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

The work done by gravity is given by the formula,

W = mgh

W = 50 x 9.8 x ( 20-8)

= 5880 J

So the work done is 5880 J

Learn more about Work done here:

brainly.com/question/25239010

#SPJ4

7 0
2 years ago
A physics instructor wants to project a spectrum of visible-light colors from 400 nm to 700 nm as part of a classroom demonstrat
Likurg_2 [28]

Answer:

  Dr = 263 10⁻⁶ m

Explanation:

The diffraction pattern for constructive interference is described by

        a sin θ = m λ

in this it indicates that the order of diffraction is m = 1

Let's use a direct proportion rule to find the separation of two slits. If there are 600 lines in 1 me, what is the distance between 2 slits

   a = 2 lines 1/600

   a = 2/600

    a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm

let's use trigonometry

      tan θ = y / L

as the measured angles are small

      tan θ = sin θ / cos θ sin θ

      sin θ = y / L

we substitute

     a  y/L = λ

     y = λ L / a

for λ = 400 10-9 m

      I = 400 10⁻⁹ 2.9 / 3.33 10⁻³

      i = 346.89 10⁻⁶ m

f

or λ = 700 nm

        y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³

        y_f = 609.609 10⁻⁶ m

the separation of this spectrum

        Δr = v_f - i

        Dr = (609.609 - 346)  10 ⁻⁶

        Dr = 263 10⁻⁶ m

4 0
3 years ago
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