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Veronika [31]
3 years ago
11

If you ran 15 km/hr for 2.5 hours, how much distance would you cover?

Physics
1 answer:
SVETLANKA909090 [29]3 years ago
6 0

Answer: 37.5 km

Explanation:

The question is that

If you ran 15 km/hr for 2.5 hours, how much distance would you cover ?

Where

Speed = 15 km/ hr

Time = 2.5 hours

Using the formula for speed.

Speed = distance/time

Substitute speed and time into the formula

15 = distance/ 2.5

Make distance the subject of formula by cross multiplying.

Distance = 15 × 2.5

Distance = 37.5 km.

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A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the
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Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

\frac {Vf^{2}-Vo^2}{2.a} =X

Where

Vf = final speed

Vo = Initial speed =7.3m/S

A = g=acceleration =9.81m/s^2

X = displacement =51m}

solving for Vf

Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s

the speed with the ball hits the ground is 32.46m/s

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How to charge a laptop battery without a charger?
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Lindsay is late to class and is running down the hallway at a speed of 2 m/s the length of the hallway is 150 meters . How long
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Explanation:

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This is a question on my physics test :)
Licemer1 [7]

Answer:

119.6 J/Kg°C

Explanation:

Data obtained from the question include:

Mass of substance (ms) = 170 g

Initial temperature of substance (Ts) = 120 °C

Volume of water = 200 mL

Initial temperature of water (Ts) = 10 °C

Temperature of the mixture (T2) = 12.6 °C

Density of water = 1 g/mL

Specific heat capacity of water (Cw) = 4200J/Kg°C

Specific heat capacity of substance (Cs) =..?

Next, we shall determine the mass of water. This can be obtained as follow:

Volume of water = 200 mL

Density of water = 1 g/mL

Mass of water =..?

Density = mass /volume

1 = mass /200

Cross multiply

Mass of water = 1 x 200

Mass of water = 200 g

Convert 200 g of water to Kg

Mass of water = 200/1000 0.2 Kg

Mass of water = 0.2 Kg

Now, we obtained the specific heat capacity of the substance using the following formula:

MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0

Mass of water = 0.2 Kg

Initial temperature of water (Ts) = 10 °C

Specific heat capacity of water (Cw) = 4200J/Kg°C

Temperature of the mixture (T2) = 12.6 °C

Mass of substance (ms) = 170 g = 170/1000 = 0.17 Kg

Initial temperature of substance (Ts) = 120 °C

Specific heat capacity of substance (Cs) =..?

MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0

0.2× 4200(12.6 – 10) + 0.17×Cs×(12.6 – 120) = 0

840(2.6) + 0.17Cs(– 107.4) = 0

2184 – 18.258Cs = 0

Rearrange

2184 = 18.258Cs

Divide both side by the coefficient of Cs i.e 18258

Cs = 2184/18.258

Cs = 119.6 J/Kg°C

Therefore, the specific heat capacity of the substance is 119.6 J/Kg°C

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