Answer: the average speed of the rat from the information given above is 0.7m/s
Explanation:
position is given as
x(t) = pt² + qt
finding the diffencial of x(t) with respect to t, we have
d(x(t))/dt = 2pt + q
we substitute the p = 0.36m/s² and q= -1.10 m/s
d(x(t))/dt = 2(0.36)t + (-1.10)
so, at t= 1s
d(x(t))/dt = 2*(0.36)*1 - 1.1 = 0.72 - 1.1 = -0.38m/s
at t= 4s
d(x(t))/dt = 2*(0.36)*4 - 1.10 = 2.88 - 1.10 = 1.78 m/s
To find the average speed,
average speed = (V1 + V2)/ 2
average speed = (1.78 + (-0.38))/2 = 0.7m/s
The impulse given to the ball is equal to the change in its momentum:
J = ∆p = (0.50 kg) (5.6 m/s - 0) = 2.8 kg•m/s
This is also equal to the product of the average force and the time interval ∆t :
J = F(ave) ∆t
so that if F(ave) = 200 N, then
∆t = J / F(ave) = (2.8 kg•m/s) / (200 N) = 0.014 s
Answer:
Vd = 1.597 ×10⁻⁴ m/s
Explanation:
Given: A = 3.90×10⁻⁶ m², I = 6.00 A, ρ = 2.70 g/cm³
To find:
Drift Velocity Vd=?
Solution:
the formula is Vd = I/nqA (n is the number of charge per unit volume)
n = No. of electron in a mole ( Avogadro's No.) / Volume
Volume = Molar mass / density ( molar mass of Al =27 g)
V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³
n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)
n= 6.02 × 10 ²⁸
Now
Vd = (6A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.9×10⁻⁶ m²)
Vd = 1.597 ×10⁻⁴ m/s
Answer:
The direction of the B-field is in the +y-direction.
Explanation:
The corresponding formula is

This means, we should use right-hand rule.
Our index finger is pointed towards +x-direction (direction of velocity),
our middle finger should point towards the direction of the B-field,
and our thumb should point towards the +z-direction (direction of the force).
Since our middle finger in this situation points towards +y-direction, the B-field should be in +y-direction.

If a particle undergoes simple harmonic motion with an amplitude of 0.21 meters, this means that the maximum displacement of the particle from its resting position is 0.21. For one period, it traveled from its starting position which is twice the amplitude and then back to its original position which is another distance that is twice the amplitude as well. Therefore, the total distance it traveled is 2*amplitude + 2*amplitude = 2*0.21 + 2*0.21 = 0.42 + 0.42 = 0.84 meters.