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3 years ago

What is the Kinetic Energy of a 60 kg person on skis traveling 20 m/s?

Physics

1 answer:

sattari [20]3 years ago

3 0

Explanation:

equating the parameters into the formula, it's gonna be

= ½ × 60 × 20²

= ½ × 60 × 400

= ½ × 24000

K.E = 12000J

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A 0.5 kg ball moves in a circle that is 0.4 m in radius at a speed of 4.0 m/s. Calculate the force exerted on the ball.

blsea [12.9K]

Answer:

Explanation:

Given a ball of mass m= 0.5kg

The ball moves in as circle of radius r= 0.4m

Speed of the ball is v = 4m/s

Centripetal force is exerted on ball and it is given as

Fc = m•ac

ac is centripetal acceleration and it is given as

ac = v²/r

Then,

Fc = mv²/r

Fc = 0.5 × 4²/0.4

Fc = 20N.

The force exerted on the ball is 20N

5 0

3 years ago

If you have a 1.0 m aqueous solution of NaCl, by how much will it increase the water’s boiling point, if KB = 0.512 °C/m? In oth

fgiga [73]

<u>Answer:</u> The elevation in boiling point is 1.024°C.

<u>Explanation:</u>

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=ik_b\times m$

where,

i = Van't Hoff factor = 2 (for NaCl)

$\Delta T_b$ = change in boiling point = ?

$k_b$ = boiling point constant = $0.512^oC/m$

m = molality = 1.0 m

Putting values in above equation, we get:

$\Delta Tb=2\times 0.512^oC/m\times 1.0m\\\\\Delta Tb=1.024^oC$

Hence, the elevation in boiling point is 1.024°C.

5 0

3 years ago

a car slows down from 22 m/s to 3 m/s at a constant rate of 3.5 m/s squared. how many seconds are required before the car is tra

Stolb23 [73]

Answer:

9.05 s

Explanation

v=u+at

6 0

3 years ago

Select the statement that is not true about the Hubble Telescope:

xxMikexx [17]

Answer:

The option is B is not true for Hubble telescope.

4 0

3 years ago

An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 5.00 V. T

Maksim231197 [3]

Answer:

Explanation:

We are given that

Initial voltage, $V_1=120 V$

Final voltage, $V_2=5 V$

Number of tuns in primary coil of the transformer, $N_p=840$

Rms current, $I_{rms}=580mA=580\times 10^{-3} A$

$1 mA=10^{-3} A$

We have to find the number of turns are there on the secondary coil.

We know that

$\frac{N_s}{N_p}=\frac{V_2}{V_1}$

Using the formula

$\frac{N_s}{840}=\frac{5}{120}$

$N_s=\frac{5}{120}\times 840=35$

Hence, there are number of turns on the secondary coil=35

8 0

3 years ago