Answer:
250,000
Explanation:
<h2> </h2>
<h2>formula = ( F=ma </h2>
- F=1500N
- a=6m/s^2
- F= ma
- m=?
- 1500/6 = m
- m=250 kg
- 1kg =1000gm so 250kg =250,000gm
- m =250×10^3 gm
<span>Like charges repel and opposite charges attract.
The further away two charged objects are the weaker the electrical force between them.
The closer two charged objects are the stronger the electrical force between them.
Hope this helps :)</span>
Answer:
θ = 22.2
Explanation:
This is a diffraction exercise
a sin θ = m λ
The extension of the third zero is requested (m = 3)
They indicate the wavelength λ = 630 nm = 630 10⁻⁹ m and the width of the slit a = 5 10⁻⁶ m
sin θ = m λ / a
sin θ = 3 630 10⁻⁹ / 5 10⁻⁶
sin θ = 3.78 10⁻¹ = 0.378
θ = sin⁻¹ 0.378
to better see the result let's find the angle in radians
θ = 0.3876 rad
let's reduce to degrees
θ = 0.3876 rad (180º /π rad)
θ = 22.2º
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.
Observer A is moving inside the train
so here observer A will not be able to see the change in position of train as he is standing in the same reference frame
So here as per observer A the train will remain at rest and its not moving at all
Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body
So here observer B will see the actual motion of train which is moving in forward direction away from the platform
Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction
So the distance between them will decrease at faster rate
Now as per Newton's II law
F = ma
Now if train apply the brakes the net force on it will be opposite to its motion
So we can say
- F = ma
so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate
It is not affected by the gravity because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train
So there is no effect on train motion
