If a galaxy is located 200 million light years from Earth, what can you conclude about the light from that galaxy?

If a 42kg rolling object slows from 11.5 m/s to 3.33 m/s how much work does friction do?

lara31 [8.8K]

Explanation:

work done by friction = 1/2 x 42 x ( 3.33^2 - 11.5^2)

= 21 ( 11.08 - 132.25)

= 21 ( - 121.17 )

= - 2544.57 J

4 0

3 years ago

When light traveling in straight lines passes through an object that is curved like a lens, the light is ________ at different a

qwelly [4]

Answer:

You are right. It is reflected

Explanation:

if it is concave then the light is reflected at one direction.

if it is convex the light is reflected outwards and they will travel away from the focal point and the light scatters or will be reflected to different directions

3 0

2 years ago

Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0

3 years ago

A 10.0g piece of copper wire, sitting in the sun reaches a temperature of 80.0 C. how many Joules are released when the copper c

Zolol [24]

Answer:

150.8 J

Explanation:

The heat released by the copper wire is given by:

$Q=mC_s \Delta T$

where:

m = 10.0 g is the mass of the wire

Cs = 0.377 j/(g.C) is the specific heat capacity of copper

$\Delta T=40.0 C - 80.0 C=-40.0 C$ is the change in temperature of the wire

Substituting into the equation, we find

$Q=(10.0 g)(0.377 J/gC)(-40.0^{\circ})=-150.8 J$

And the sign is negative because the heat is released by the wire.

6 0

3 years ago

A car is moving with an initial relocity of

MA_775_DIABLO [31]

Answer:

The final acceleration of the car, v = 70 m/s

Explanation:

Given,

The initial velocity of the car, u = 20 m/s

The acceleration of the car, a = 10 m/s²

The time period of travel, t = 5 s

Using the I equations of motion

v = u + at

= 20 + 10(5)

= 20 + 50

= 70 m/s

Hence, the final acceleration of the car, v = 70 m/s

4 0

2 years ago