Answer:
35 mph
Explanation:
From the question given above, the following data were obtained:
Time (t) = 2 hours
Distance (d) = 70 miles
Speed (S) =.?
Speed is simply defined as the distance travelled per unit time. Mathematically, it is expressed as:
Speed (S) = Distance (d) / Time (t)
S = d/t
With the above formula, we can calculate how fast (i.e the speed) Joe will need to drive to make it to his Grandma's house on time. This can be obtained as follow:
Time (t) = 2 hours
Distance (d) = 70 miles
Speed (S) =.?
S = d/t
S = 70/2
S = 35 mph
Therefore, Joe will drive at 35 mph in order to get to his Grandma's house on time.
Magnetic field B is produced when a current I Amphere passes through a solenoid. B is parallel to its axis.
B=U N/I I. N is number of turns in the solenoid of lm length
N=200, l= 20cm= 0.2m, I = 1₀sin (2πft) where f is equal to 60Hz
B= 4π × 10⁻⁷(200/0.2) l₀ sin (2πft) T
=1.256 × 10⁻³ l₀ sin(2πft) Tesla
Area of the coil is πr² = π (1.5cm)² = 2.25π ×10⁴m²
magnetic flux which is through the coil is given by
Ф = B.A = BA cosФ
ФФ = O since B is in direction of A
A= 40 ×π×2.25 ×⁻⁴m² which is the number of turns being 40.
Flux Ф through the coil is,
1.256 ×10⁻³ l₀ sin (2πft) ×9π ××10⁻³m²
=35.5 ×10⁻⁶ l₀ sin ₀(2πft)ab
Ф is time-varying emf will be generated in the coil
∈= dФ/dt
∈ = d/dt [35.5 × 10⁻d l₀ sin (2πft) ab]
∈ = 35.5 ×10⁻⁶ l₀ 2πf cos 2πftV
f = 60Hz
∈∈ 13376.4 ×10⁻⁶ l₀ cos 2πftV
Current I amp shall be induced in the cell of resistance Rohm so
I= E/R
I = 13376.4 ×10⁻⁶ l₀ cos 2πft)V/0.4∩
=33441 ×10⁻⁶ I₀ cos 2πft A
I = 3344q × 10 ⁻⁶ l₀
But I = 0.2A
l₀ = (0.2)(10⁶)/33441 = 6.0A
Answer:
The sun is pulled toward the moon.
Explanation:
It is the sun is pulled toward the moon because the moon doesn't have a great enough gravitational force or mass to pull the sun, so it must be some external force.
Answer:
The work done on the canister is 15.34 J.
Explanation:
Given;
mass of canister, m = 1.9 kg
magnitude of force acting on x-y plane, F = 3.9 N
initial velocity of canister in positive x direction, = 3.9 m/s
final velocity of the canister in positive y direction,
The change in the kinetic energy of the canister is equal to net work done on the canister by 3.9 N.
ΔK.E =
ΔK.E
The initial kinetic energy of the canister;
The final kinetic energy of the canister;
ΔK.E = 29.79 J - 14.45 J
ΔK.E = = 15.34 J
Therefore, the work done on the canister is 15.34 J.