Answer:
The image is formed at 0.44 m in front of the mirror
magnification (M) = 0.44
Explanation:
Applying, mirror formular
1/f = (1/u)+(1/v).................... Equation 1
Where f = Focal length of the convex mirror, u = object distance, v = image distance.
Using the real is positive convection,
From the question,
Given: f = -1.6/2 = -0.8 m( The focal length of a convex mirror is vitual), u = 1.0 m
Substitute these values into equation 1
-1/0.8 = (1/v)+(1/1)
Solve for v
1/v = 1.25+1
1/v = 2.25
v = 1/2.25
v = 0.44 m
Hence the image is formed at 0.44 m in front of the mirror
Magnification (M) = v/u
m = 0.44/1
m = 0.44
define Neutons 3rd law .
Answer:
that when two object interact, they apply forces to one another that are equal in magnitude and opposite in direction.
Explanation:
Newton's third law: the law of action and reaction
What do is mean by environment friendly behaviour?
derive an expression for resistivity of conductor of length l and area of cross section A
Answer
Resistivity R = K * L / A where resistivity is constant for material, L the length of the material and A the area of the material
K = R * A / L where R is the resistance of the material
Saul is testing an installation and discovers a short circuit what’s causing this
A.high current
B.high voltage
C.low resistance
D.low voltage
Answer:
im pretty sure that it's (A.) High current
Explanation:
pf
not 100% sure, tho
the slope of a line on a position vs time graph is the
a. velocity
b. time
c. distance
d. displacement
It's the velocity, but only the magnitude. It can't show the direction of the velocity. So it's better to call it speed.
Given: F = k· m. g
Solve for "k"
Answer:
[tex]F = kmg \\ k = \frac{F}{mg} [/tex]
Explanation:
F = k . m . g
=> F = k . mg
[tex] = > k = \frac{F}{mg} (ans)[/tex]
why Fossil fuel has been used more in the existing world ?
A car has a mass of 2000 kg. While it is traveling along a perfectly flat road, it goes around an unbanked turn that has a radius of 40.0 m. The coefficient of static friction between the car tires and the road is 0.500. The car travels successfully around the turn at a constant speed of 10.0 m/s. Calculate the magnitude of the car's acceleration as it goes around the turn. _______ m/s^2
Answer:
2.5 m/s²
Explanation:
The given parameters are;
The mass of the car, m = 2,000 kg
The radius of the car, r = 40.0 m
The coefficient of friction between the car tires and the road, μ = 0.500
The constant speed with which the car moves, v = 10.0 m/s
The normal reaction of the road on the car, N = The weight of the car;
∴ N = m × g
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
N ≈ 2,000 kg × 9.81 m/s² = 19,620 N
The frictional force, [tex]F_f[/tex] = μ × N
The centripetal force, [tex]F_c[/tex] = m·v²/r
The car moves without slipping when [tex]F_f[/tex] = [tex]F_c[/tex]
Therefore, [tex]F_f[/tex] = 0.500 × 19,620 N = 2,000 kg × [tex]v_{max}[/tex]²/40.0 m
∴ [tex]v_{max}[/tex] = √(0.500 × 19,620 N × 40.0 m/2,000 kg) ≈ 14.007 m/s
Therefore, the velocity with which the car moves, v < [tex]v_{max}[/tex]
The cars centripetal acceleration, [tex]a_c[/tex] = v²/r
∴ [tex]a_c[/tex] = (10.0 m/s)²/40.0 m = 2.5 m/s²
The cars centripetal acceleration as it goes round the turn, [tex]a_c[/tex] = 2.5 m/s².
