Answer:
a=3.53 m/s^2
Explanation:
Vo=0 m/s (because he is not moving at the start)
V1=15 m/s
t= 4.25 s
a = (V1-Vo) / t = 15/4.25 = 3.53 m/s^2
Answer:
0.546 ohm / μm
Explanation:
Given that :
N = 1.015 * 10^17
Electron mobility, u = 3900
Hole mobility, h = 1900
Ng = 4.42 x10^22
q = 1.6*10^-19
Resistivity = 1/qNu
Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)
= 0.01578880889 ohm /cm
Resistivity of germanium :
R = 1 / 2q * sqrt(Ng) * sqrt(u*h)
R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)
R = 1 /0.0001831
R = 5461.4964 ohm /cm
5461.4964 / 10000
0.546 ohm / μm
The time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds
<h3>How to calculate the time</h3>
Use the formula:
Power × time = mass × specific heat
Given mass = 1. 5kg
Specific latent heat of vaporization = 4000000 J/ Kg
Power = 2500 W
Substitute the values into the formula
Power × time = mass × specific heat
2500 × time = 1. 5 × 4000000
Make 'time' the subject
time = 1. 5 × 4000000 ÷ 2500 = 6000000 ÷ 2500 = 2400 seconds
Therefore, the time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds.
Learn more about specific latent heat of vaporization:
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<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
