The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23
1 answer:
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Answer:
The magnitude of the torque on the loop due to the magnetic field is .
Explanation:
Given that,
Diameter = 10 cm
Current = 0.20 A
Magnetic field = 0.30 T
Unit vector
We need to calculate the torque on the loop
Using formula of torque
Where, N = number of turns
A = area
I = current
B = magnetic field
Put the value into the formula
Hence, The magnitude of the torque on the loop due to the magnetic field is .
Answer:
Answer:
1.1 x 10^9 ohm metre
Explanation:
diameter = 1.5 mm
length, l = 5 cm
Potential difference, V = 9 V
current, i = 230 micro Ampere = 230 x 10^-6 A
radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m
Let the resistivity is ρ.
Area of crossection
A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2
Use Ohm's law to find the value of resistance
V = i x R
9 = 230 x 10^-6 x R
R = 39130.4 ohm
Use the formula for the resistance
ρ = 1.1 x 10^9 ohm metre
Explanation:
Answer:
15.1°
Explanation:
The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:
Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:
(1)
where
is the initial vertical velocity
g = 9.8 m/s^2 is the gravitational acceleration
t is the time
Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by
Substituting t into (1) we find the final vertical velocity
where the negative sign means that the velocity is downward.
Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal: