Answer:
Magnitude of the induced emf is 11.62 V
Explanation:
Given:
No. of turns of the circular coil, N=185
Radius, R=4.50 cm=0.045 m
No. of turns of solenoid per meter(m), n=350
a=12.5 A
b=2.10
Now,
To determine the emf induced in the coil at t = 1.50 s:
The given equation is:
Now,
Now, substituting the respective values:
Now,
where,
A=Area=
Thus,
I think it would be negative apex :)
Acceleration units are in m/s^2 , so you take 25.5 m/s divided by 5.75s. Acceleration is the rate of change of velocity :)
follow me on instagram : imrajsingh :):) thanks
Looking out from a vertex with angle θ, sin(θ) is the ratio of the opposite side to the hypotenuse , while cos(θ) is the ratio of the adjacent side to the hypotenuse . No matter the size of the triangle, the values of sin(θ) and cos(θ) are the same for a given θ, as illustrated below.
<h2>Mark as brainlest answer!</h2>
Answer:
1010 m
Explanation:
The following data were obtained from the question:
Height (h) of cliff = 500 m
Horizontal velocity (u) = 100 m/s
Horizontal distance (s) =?
Next we shall determine the time taken for the cannon ball to hit the ground. This can be obtained as follow:
Height (h) of cliff = 500 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
500 = ½ × 9.8 × t²
500 = 4.9 × t²
Divide both side by 4.9
t² = 500/4.9
Take the square root of both side
t = √(500/4.9)
t = 10.1 s
Finally, we shall determine the horizontal distance travelled by the cannon ball.
This can be obtained by using the following formula (s = ut) as illustrated below:
Horizontal velocity (u) = 100 m/s
Time (t) = 10.1 s
Horizontal distance (s) =?
s = ut
s = 100 × 10.1
s = 1010 m
Thus, the cannon ball was launched 1010 m away from the cliff.