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Korvikt [17]
3 years ago
11

According to the law of reflection, what is the angle of incidence

Physics
2 answers:
eimsori [14]3 years ago
4 0

Answer:

It is the angle the incident ray makes with a line drawn perpendicular to the viewer.

Explanation:

gradpoint

Sergio [31]3 years ago
3 0

Answer:

It is the angle the incident ray makes with a line drawn perpendicular to the viewer.

Explanation:

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In which situation are both the speed and velocity of the car changing?
lubasha [3.4K]
Answer is A because the speed and velocity would change. Think of it as GTA, your going 30+ miles per hour and you take a left turn, the speed and velocity would change in an instant.. 
Hope this helped.
6 0
3 years ago
Read 2 more answers
An empty parallel plate capacitor is connected between the terminals of a 18.8-V battery and charges up. The capacitor is then d
Basile [38]

Answer:

p.d' = 37.6 V

Explanation:

From the question we are told that:

Potential difference p.d=18.8V

New Capacitor C_1=C_2/2

Generally the equation for Capacitor capacitance is mathematically given by

C=\frac{eA}{d}

Generally the equation for New p.d' is mathematically given by

 C_2V=C_1*p.d'

  p.d' = 2V

 p.d'= 2 * 18.8

 p.d' = 37.6 V

7 0
3 years ago
Suppose astronomers find an earthlike planet that is twice the size of Earth (that is, its radius is twice the radius of Earth).
JulijaS [17]

Answer:

4 times the mass of Earth

Explanation:

M_1 = Mass of Earth

M_2 = Mass of the other planet

r = Radius of Earth

2r = Radius of the other planet

m = Mass of object

The force of gravity on an object on Earth is

F=\frac{GM_1m}{r^2}

The force of gravity on an object on the other planet is

F=\frac{GM_2m}{(2r)^2}

As the forces are equal

\frac{GM_1m}{r^2}=\frac{GM_2m}{(2r)^2}\\\Rightarrow M_1=\frac{M_2}{4}\\\Rightarrow M_2=4M_1

So, the other planet would have 4 times the mass of Earth

6 0
3 years ago
The space shuttle orbits 310 km above the surface of the Earth.
MAVERICK [17]

Answer:

44.7 N

Explanation:

The gravitational force between the objects is given by:

F=G\frac{mM}{r^2}

where

G is the gravitational constant

m and M are the masses of the two objects

r is the distance between the centres of the two objects

In this problem, we have:

m=5.0 kg is the mass of the sphere

M=5.98\cdot 10^{24} kg is the Earth's mass

R=6370 km is the Earth's radius, while h=310 km is the altitude of the sphere, so the distance of the sphere from Earth's centre is

r=6370 km+310 km=6680 km=6.68\cdot 10^6 m

Substituting into the equation, we find

F=(6.67\cdot 10^{-11})\frac{(5.0 kg)(5.98\cdot 10^{24} kg)}{(6.68\cdot 10^6 m)^2}=44.7 N

8 0
3 years ago
Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A
kolezko [41]

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × \frac{ \lambda }{2}      ........1

so  λ = \frac{2L}{10}  

and velocity is express as

V = \sqrt{\frac{T}{\mu } }    .................2

so

frequency for string A = f1 = \frac{V1}{\lambda}

f1 = \frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}

f1 = \frac{10}{2L} \sqrt{\frac{T1}{\mu } }      

and

f2 = \frac{10}{2L} \sqrt{\frac{T2}{\mu } }

so

beat frequency is = f2 - f1

put here value

beat frequency = \frac{10}{2*2} \sqrt{\frac{130}{0.0065}}  - \frac{10}{2*2} \sqrt{\frac{120}{0.0065} }

beat frequency = 13.87 Hz

6 0
3 years ago
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