We can use the kinematic equation
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance
we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s [if you are rounding this by significant figures]
Answer:
The magnitude of the applied torque is 
(e) is correct option.
Explanation:
Given that,
Mass of object = 3 kg
Radius of gyration = 0.2 m
Angular acceleration = 0.5 rad/s²
We need to calculate the applied torque
Using formula of torque
Here, I = mk²
Put the value into the formula
Hence, The magnitude of the applied torque is
S = ut + 0.5at^2
<span>10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} </span>
<span>t^2 ~ 2.04 </span>
<span>t ~ 1.43 seconds</span>
Answer:
C
Explanation:
horizintal speed stays same
only vertical speed changes
so H speed will stay 30 m/s
Answer:
(a) 
(b) 
Explanation:
Parameter given:
Electric field, E = 
(a) Electric force is given (in terms of electric field) as a product of electric charge and electric field.
Mathematically:
Electric charge, q, of an electron = 
(b) This electrostatic force causes the electron to accelerate with an equivalent force:
F = -ma
where m = mass of an electron
a = acceleration of electron
(Note: the force is negative cos the direction of the force is opposite the direction of the electron)
Therefore:
Mass, m, of an electron = 
=> 
The acceleration of the electron is 
