Answer:
6.71 × 10^8 mi/hr
Explanation:
Light is usually defined as an electromagnetic wave that is comprised of a definite wavelength. It is of both types, visible and invisible. The light emitted from a source usually travels at a speed of about 3 × 10^8 meter/sec. This speed of light is commonly represented by the letter 'C'.
To write it in the metric system, it has to be converted into miles/hour.
We know that,
1 minute = 60 seconds
60 minutes = 1 hour
1 kilometer = 1000 meter
1 miles = 1.6 kilometer
Now,
= 
= 1.08 × 10^12 m/ hr (meter/hour)
= 
= 6.71 × 10^8 mi/hr (miles/hour)
Thus, the value for speed of light (C) in metric unit is 6.71 × 10^8 mi/hr.
complete question:
A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
F = 1776 N
Explanation:
mass of ball = 60 g = 0.06 kg
velocity of downward direction = 22 m/s = v1
velocity of upward direction = 15 m/s = v2
Δt = 1/800 = 0.00125 s
Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.
p = mv
When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .
I = pf − pi = ∆p
F = ∆p/∆t = I/∆t
let the upward velocity be the positive
Δp = mv2 - m(-v1)
Δp = mv2 - m(-v1)
Δp = m (v2 + v1)
Δp = 0.06( 15 + 22)
Δp = 0.06(37)
Δp = 2.22 kg m/s
∆t = 0.00125
F = ∆p/∆t
F = 2.22/0.00125
F = 1776 N
Answer:9.34 A/s
Explanation:
Given
radius of solenoid 
Emf induced 
no of turns per meter n=450
we know Induced EMF is given by
Magnetic Field is given by
thus 
Area of cross-section
where
solving integration we get
where r=distance from axis
R=radius of Solenoid
what causes a star to shine brightly:
by squeezing atoms together in its core
Given :
An object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m.
To Find :
the image height (in cm).
Solution :
By lens formula :
Here, u = - 100 cm
f = 150 cm
Now, magnification is given by :
Therefore, the image height is 3 m or 300 cm.
