Answer:
m = 0.25
Explanation:
Given that,
Object distance, u = -15cm
Height of the object, h = 48
Focal length, f = cm
We need to find the magnification of the image.
Let v is the image distance. Using mirror's equation.
Magnification,
Hence, the magnification of the image is 0.25.
20h
Explanation:
Given parameters:
Electrical power = 75W
Electrical energy = 1500J
Unknown:
Time taken = ?
Solution:
Electrical power is the amount of voltage that drives current.
P = IV
Electrical energy is the voltage that drives a current per unit of time
E = IVt
E = Pt
Since time is the unknown, we can make it subject of the formula;
t = =
= 20h
Learn more:
Invention of electrical power brainly.com/question/13146914
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Answer:
The angular displacement is
Explanation:
From the question we are told that
The initial angular speed is
The angular acceleration is
The time take is
Generally the angular displacement is mathematically represented as
substituting values
Answer:
The ball lands 154.3 ft from the origin at an angle of 13.6° from the eastern direction toward the south.
Explanation:
Hi there!
The position vector of the ball is described by the following equation:
r = (x0 + v0x · t + 1/2 · ax · t², y0 + v0y · t + 1/2 · ay · t², z0 + v0z · t + 1/2 · g · t²)
Where:
r = poisition vector of the ball at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity (eastward).
t = time.
ax = horizontal acceleration (eastward).
y0 = initial horizontal position.
v0y = initial horizontal velocity (southward).
ay = horizontal acceleration (southward)
z0 = initial vertical position.
v0z = initial vertical velocity.
g = acceleration due to gravity.
We have to find at which time the vertical component of the position vector is zero (the ball is on the ground) and then we can calculate the horizontal distance traveled by the ball at that time, using the equations of the horizontal components of the position vector.
Let´s place the origin of the system of reference at the throwing point so that x0 and y0 and z0 = 0.
y = z0 + v0z · t + 1/2 · g · t² (z0 = 0)
0 = 48 ft/s · t - 1/2 · 32 ft/s² · t²
0 = t (48 ft/s - 16 ft / s² · t) (t= 0, the origin point)
0 = 48 ft/s - 16 ft / s² · t
- 48 ft/s / -16 f/s² = t
t = 3.0 s
Now, we can calculate how much distance the ball traveled in that time.
First, let´s calculate the distance traveled in the eastward direction:
x = x0 + v0x · t + 1/2 · ax · t² (x0 = 0, ax = 0 there is no eastward acceleration)
x = 50 ft/s · 3 s
x = 150 ft
And now let´s calculate the distance traveled in southward direction:
y = y0 + v0y · t + 1/2 · ay · t² (y0 = 0 and v0y = 0, initially, the ball does not have a southward velocity).
y = 1/2 · ay · t²
y = 1/2 · (-8 ft/s²) · (3 s)²
y = -36 ft
Then, the final position vector will be:
r = (150 ft, -36 ft, 0)
The traveled distance is the magnitude of the position vector:
To calculate the angle, we have to use trigonometry (see attached figure):
cos angle = adjacent side / hypotenuse
cos α = x/r
cos α = 150 ft / 154.3 ft
α = 13.6°
The ball lands 154.3 ft from the origin at an angle of 13.5° from the eastern direction toward the south.
