Answer: 15.8
Explanation:
You are given that the
Object distance U = 32 cm
Focal length F = 30.1 cm
First calculate the image distance V by using the formula
1/F = 1/U + 1/V
Substitute F and V into the formula
1/30.1 = 1/32 + 1/V
1/V = 1/30.1 - 1/32
1/V = 0.00197259
Reciprocate both sides
V = 506.94 cm
Magnification M is the ratio of image distance to object distance.
M = V/U
substitute the values of V and U into the formula
M = 506.94/32
M = 15.8
Therefore, the magnification of the image is 15.8 or approximately 16.
Answer:
A
Explanation:
From a Solenoid we know that a magnetic fiel is always inversely proportional to lenght L or BL = constant
As I is constant
Let m = mass of asteroid y.
Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.
Given:
F = 6.2x10⁸ N
d = 2100 km = 2.1x10⁶ m
Note that
G = 6.67408x10⁻¹¹ m³/(kg-s²)
The gravitational force between the asteroids is
F = (G*m*(m/3))/d² = (Gm²)/(3d²)
or
m² = (3Fd²)/G
= [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))
= 1.229x10³² kg²
m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)
Answer: 1.1x10¹⁶ kg
GIVEN:
60 beats per minute
21 beats per minute
find x= how fast would an astronaut be flying away
1 x
----- * ------ = (60x = 21) -------> 60x = 21 ------------> x= 0.35
60 21 ------- -----
60 60
The answer is 0.35 seconds which refers to how fast would an astronaut be flying away from the earth if he has a heart rate of 21 beats/min.
Answer:
The value of the linear coefficient of thermal expansion is : α=1.01 *10⁻⁵ (ºC)⁻¹
Explanation:
Li = 0.2m
ΔL = 0.2 mm = 0.0002m
T1 = 21ºC
T2 = 120ºC
ΔT =99ºC
α =ΔL/(Li*ΔT)
α =0.0002m /(0.2m * 99ºC)
α = 1.01 *10⁻⁵ (ºC)⁻¹
