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ololo11 [35]
3 years ago
5

A 16-kg scooter is moving at a speed of 7 m/s. The scooter’s speed doubles. What is the scooter’s kinetic energy when its speed

doubles?
a. 784 J
b. 1568 J
c. 392 J
d. 112 J
Physics
1 answer:
lbvjy [14]3 years ago
5 0
The formula for kinetic energy is:

KE = 0.5mv^2

where:
m = mass
v = speed

Given this formula, the original KE of the scooter is calculated as 392 J. Since the speed is doubled, it can be observed in the formula that the change would actually affect the kinetic energy by quadrupling it. This is because 2^2 is 4. So, 392(4) = 1568 J or B.
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Why does the value of g vary from place to place?
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8 0
2 years ago
Please help answer question​
nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, v_t=60\ m/s

Area of cross section, A=0.33\ m^2

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

or

\dfrac{1}{2}\rho CAv_t^2=mg

Where

\rho is the density of air = 1.225 kg/m³

C is drag coefficient

So,

C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01

So, the drag coefficient is 1.01.

4 0
3 years ago
A 460 W heating unit is designed to operate with an applied potential difference of 120 V (a) By what percentage will its heat o
dybincka [34]

Answer:

(a) = -0.16%

(b) = smaller

Explanation:

given

power = 460 W

potential difference = 120 V

(a) what percentage will   its heat output drop if the applied potential difference drops to 110 V ?

we know p = \frac{v^2}{R} .....................(i)

we need to find change in power

\Delta P = \frac{\Delta (V^2)}{R}  

\Delta P = \frac{2 V \Delta V}{R}..............(ii)

from equations we get

\frac{\Delta P}{P} =  \frac{2 \Delta V}{V}

\frac{\Delta P}{P} = 2 \frac{110 -120}{120}

\frac{\Delta P}{P} =  -2(\frac{10}{120})

\frac{\Delta P}{P} = - 0.16 %

(b)

if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power

and due to this increment of heating power resistance will decrease so actual drop in the power would  be smaller

7 0
3 years ago
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