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agasfer [191]
3 years ago
9

A ball is thrown directly downward with an initial speed of 8.75 m/s, from a height of 21.0 m. after what interval does the ball

strike the ground

Physics
1 answer:
gayaneshka [121]3 years ago
8 0
There you go!
as it is unnecessary to find velocity, so use the equation with v
remember to take all directions downwards as it is more convenient to calculate
after substituting the values, you have to reject the negative answed as time must be+
:)

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How much potential energy does a 50-N box have when lifted at a height of 1.5M?
nikitadnepr [17]

The correct answer is: Option (A) 75 J

Explanation:

First, be careful with the units here. As you can see it is mentioned that there is a 50N box. It means that the weight (<em>mg</em>) of the box is given as the unit is <em>Newton</em>, not its mass (which is in kg).

As,

Potential-energy = mass * acceleration-due-to-gravity * height

PE = m*g*h --- (A)


In equation (A), mg is actually the weight of the box, which is given.

mg = 50N

h = height = 1.5m

Plug the values in equation (A):

PE = 50 * 1.5  = <em>75 J (Option A)</em>

3 0
3 years ago
An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 ∘C and rejects he
Agata [3.3K]

Answer:

2.36 x 10^6 J

Explanation:

Tc = 0°C = 273 K

TH = 22.5°C = 295.5 K

Qc = heat used to melt the ice

mass of ice, m = 85.7 Kg

Latent heat of fusion, L = 3.34 x 10^5 J/kg

Let Energy supplied is E which is equal to the work done

Qc = m x L = 85.7 x 3.34 x 10^5 =  286.24 x 10^5 J

Use the Carnot's equation

\frac{Q_{H}}{Q_{c}}=\frac{T_{H}}{T_{c}}

Q_{H}=286.24\times 10^{5}\times \frac{295.5}{273}

QH = 309.8 x 10^5 J

W = QH - Qc

W = (309.8 - 286.24) x 10^5

W = 23.56 x 10^5 J

W = 2.36 x 10^6 J

Thus, the energy supplied is 2.36 x 10^6 J.

8 0
3 years ago
Imagine that you are working as a roller coaster designer. You want to build a record breaking coaster that goes 70.0 m/s at the
Rzqust [24]

Wow !  This is not simple.  At first, it looks like there's not enough information, because we don't know the mass of the cars.  But I"m pretty sure it turns out that we don't need to know it.

At the top of the first hill, the car's potential energy is

                                  PE = (mass) x (gravity) x (height) .

At the bottom, the car's kinetic energy is

                                 KE = (1/2) (mass) (speed²) .

You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down.  So now, here comes the big jump.  Put a comment under
my answer if you don't see where I got this equation:

                                   KE = 0.9  PE

        (1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)     

Divide each side by (mass): 

               (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

(There goes the mass.  As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)

Divide each side by (0.9):

               (0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)

Divide each side by (9.8 m/s²):

               Height = (5/9)(4900 m²/s²) / (9.8 m/s²)

                          =  (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

                          =  (24,500 / 88.2)  (m²/s²) / (m/s²)

                          =        277-7/9    meters
                                  (about 911 feet)
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If a distant galaxy has a substantial redshift (as viewed from our galaxy), then anyone living in that galaxy would see a substa
mario62 [17]

Answer:

Option A

Explanation:

The statement makes sense since it's already explained that the galaxy is moving away from us and unlike option C which depicts that the galaxy is moving to us.

This statement makes sense. The redshift means that we see the galaxy moving away from us, so observers in that galaxy must also see us moving away from them—which means they see us redshifted as well

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3 years ago
Hello can someone please help me with this.
user100 [1]

Answer:

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Explanation:

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3 years ago
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