Answer:
a. F = 245 Newton.
b. Workdone = 392 Joules.
c. Power = 196 Watts
Explanation:
Given the following data;
Mass = 25kg
Distance = 1.6m
Time = 2secs
a. To find the force needed to lift the mass (in N );
Force = mass * acceleration
We know that acceleration due to gravity is equal to 9.8
F = 25*9.8
F = 245N
b. To find the work done by the student (in J);
Workdone = force * distance
Workdone = 245 * 1.6
Workdone = 392 Joules.
c. To find the power exerted by the student (in W);
Power = workdone/time
Power = 392/2
Power = 196 Watts.
The frequencies of light that an atom can emit are dependent on states the electrons can be in. When excited, an electron moves to a higher energy level or orbital. When the electron falls back to its ground level the light is emitted.
hope this helped:)
mark brainliest
Answer:
For the given conditions the fundamental frequency is 3728.26 Hertz
Explanation:
We know that for a pipe open at one end and closed at other end the fundamental frequency is given by
where
f is the fundamental frequency
is the speed of sound in air in the surrounding conditions.
L = Length of the pipe
Applying values we get and using speed of sound as 343m/s we get
Answer:
a= -0.86 m/s²
The negative sign shows that ball down the ground or moving down
Explanation:
Vf² - Vo² = 2gS
where
Vf = velocity of clay as it hits the ground
Vo = initial velocity of clay = 0
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
S = distance travelled by clay = 15 m
Substituting appropriate values,
Vf² - 0 = 2(9.8)(15)
Vf = 17.15 m/sec.
Formula to use is,
V - Vf = aT
where
V = velocity of clay when it stops = 0
Vf = 17.15 m/sec (as determined above)
a = acceleration
T = 20 ms
Put the values to find acceleration
a=(V-Vf)/T
a=(0-17.15)/20
a= -0.86 m/s²
The negative sign shows that ball down the ground
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
