Answer:
The ball will fall back and land to Elle's hands.
Explanation:
The bus move in a straight line with constant velocity means that there is no change of direction and no acceleration. Inertia can change the direction of the ball and acceleration can change its velocity. Since these two factors is not present in this scenario, the ball only has vertical movement. Thus the ball will land where it was thrown, in Elle's hands.
Answer:
The pressure of the remaining gas in the tank is 6.4 atm.
Explanation:
Given that,
Temperature T = 13+273=286 K
Pressure = 10.0 atm
We need to calculate the pressure of the remaining gas
Using equation of ideal gas
For a gas
Where, P = pressure
V = volume
T = temperature
Put the value in the equation
....(I)
When the temperature of the gas is increased
Then,
....(II)
Divided equation (I) by equation (II)
Hence, The pressure of the remaining gas in the tank is 6.4 atm.
Answer: 15.66 °
Explanation: In order to solve this proble we have to consirer the Loretz force for charge partcles moving inside a magnetic field. Thsi force is given by:
F=q v×B = qvB sin α where α is teh angle between the velocity and magnetic field vectors.
From this expression and using the given values we obtain the following:
F/(q*v*B) = sin α
3.8 * 10^-13/(1.6*10^-19*8.9*10^6* 0.96)= 0.27
then α =15.66°
Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.
