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3 years ago

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

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

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