Answer:
- 3 cm
Explanation:
From the mirror formula;
1/f = 1/v + 1/u ; where f is the focal length, v is the image distance, and u is the object distance.
1/-4.5 = 1/9 + 1/v
1/v = -1/4.5 - 1/9
= -1/3
Therefore;
v = -3 cm
Hence;
Image distance is - 3cm
Answer: 2561.7 pounds
Explanation:
If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:
(1)
Where:
is the slope of the line
is the airplane weight with 51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)
is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)
This means we already have one point of the graph, which coordinate is:

Rewritting (1):
(2)
As Y is a function of X:
(3)
Substituting the known values:
(4)
(5)
(6)
Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):
(7)
(8) This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.
I,think potential energy is mgh so 65*100*9,81
Answer:
Part 1) Voltage in secondary windings is 61.08 Volts
Part 2) Current in secondary windings is 0.53 Amperes
Explanation:
The potential developed in the primary and secondary winding of a transformer are related as

where
Np no of turns in primary coil
Ns no of turns in secondary coil
Vp Voltage of turns in primary coil
Vs Voltage of turns in secondary coil
Applying values in the formula we get

Part 2)
Using Ohm's law the current is given by

Answer:
a) t = 0.90 s, b) t = 0.815 s, c) t = 0.90 s, d) x = 3.6 m, e) t = 0.639 s
Explanation:
all these exercises are about kinematics
a) The body is released from rest,
y = y₀ + v₀ t - ½ g t²
in this case when reaching the ground y = 0 and its initial velocity is vo = 0
0 = y₀ + 0 - ½ g t²
t² = 2 y₀ / g
t² = 2 4 /9.81
t² = 0.815
t = √0.815
t = 0.90 s
b) It is thrown upwards at v₀ = 4 m / s
y = y₀ + v₀ t - ½ g t²
in this case the initial and final height is the same
y = y₀ = 0
0 = v₀ t -1/2 g t²
t = 2 v₀ / g
t = 2 4 /9.81
t = 0.815 s
c) the ball is at y₀ = 4 m and its initial velocity is horizontal v₀ = 4 m / s
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + 0 - ½ g t²
t² = 2 i / g
t² = 2 4 / 9.81
t² = 0.815
t = 0.90 s
d) the horizontal distance traveled is
x = v₀ₓ t
x = 4 0.90
x = 3.6 m
e) We can calculate the time to fall from I = 2 m
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + 0 - ½ g t²
t² = 2 y₀i / g
t² = 2 2 /9.81
t² = 0.4077
t = 0.639 s
Therefore, when making measurements, you should find readings around this value.