Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants ( = 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
= - gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
Answer:
Explanation:
Given that,
Assume number of turn is
N= 1
Radius of coil is.
r = 5cm = 0.05m
Then, Area of the surface is given as
A = πr² = π × 0.05²
A = 7.85 × 10^-3 m²
Resistance of
R = 0.20 Ω
The magnetic field is a function of time
B = 0.50exp(-20t) T
Magnitude of induce current at
t = 2s
We need to find the induced emf
This induced voltage, ε can be quantified by:
ε = −NdΦ/dt
Φ = BAcosθ, but θ = 90°, they are perpendicular
So, Φ = BA
ε = −NdΦ/dt = −N d(BA) / dt
A is a constant
ε = −NA dB/dt
Then, B = 0.50exp(-20t)
So, dB/dt = 0.5 × -20 exp(-20t)
dB/dt = -10exp(-20t)
So,
ε = −NA dB/dt
ε = −NA × -10exp(-20t)
ε = 10 × NA exp(-20t)
Now from ohms law, ε = iR
So, I = ε / R
I = 10 × NA exp(-20t) / R
Substituting the values given
I = 10×1× 7.85 ×10^-3×exp(-20×2)/0.2
I = 1.67 × 10^-18 A
Answer:
Without any external forces a moving object will continue to move in a straight line. The gravitational force between the two objects will provide the centripetal force to keep the objects moving around one another.
1. satellite in orbit around the earth (motion of earth is negligible)
2. moon in orbit around the earth (center of motion several thousand miles
from center of earth)
3. earth in orbit around sun (center of rotation close to center of sun)
4. binary stars (if masses of stars are equal center of rotation is in middle)
Answer:
Explanation:
From the given information:
radius = 15 m
Time T = 23 s
a) Speed (v) =
v = 4.10 m/s
b) The magnitude of the acceleration is:
a = 1.12 m/s²
c) True weight = mg
Apparent weight = normal force
From the top;
the normal force = upward direction,
weight is downward as well as the acceleration.
true weight - normal force = ma
apparent weight =mg - ma
= 0.886 m/s²
d)
From the bottom;
acceleration is upward, so:
apparent weight - true weight = ma
apparent weight = true weight + ma
= 1.114 m/s²