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3 years ago

5

What tells scientists about distant galaxies? a. How fast light travels c. The color of light that they give off b. If the are o

paque or transparent d. The kind of radio wave they give off

1 answer:

tamaranim1 [39]3 years ago

8 0

I Believe the answer is A but D looks good too.

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Two positive charges, 91 = 5 x 10-'[C] and q2 =1 x 10-9 [C], are

denis23 [38]

Answer:

wareffctgggyyggghhhh

4 0

3 years ago

A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated

iragen [17]

The mirror is located 30 cm from the object

Explanation:

To solve the problem, we can use the magnification equation, which states that:

$M=\frac{y'}{y}=-\frac{q}{p}$

where

M is the magnification

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

In this problem, we notice that the image formed by the mirror is erect and diminished: this means that this is a convex mirror, so the image is virtual, and this means that the image and the object are located on opposite sides of the mirror. Therefore,

$p-q=40 cm$ (distance of the image from the object is 40 cm, but since the image is virtual, q is the negative)

The size of the image is 1/3 that of the object, so

$y'=\frac{1}{3}y$

Solving the equation for p, we find the distance between the object and the mirror:

$\frac{y'}{y}=-\frac{p-40}{p}\\\\p=\frac{40}{1+\frac{y'}{y}}=\frac{40}{4/3}=30 cm$

So, the mirror is 30 cm from the object.

#LearnwithBrainly

5 0

3 years ago

It is cold and dry outside. You go down the slide and experience a small electric shock. What charge must you and slide be in or

ruslelena [56]

Answer:

c

Explanation:

the charge wants to ultimately balance out so either you or the slide gave the other electrons to balance the charges

3 0

3 years ago

Read 2 more answers

Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o

spin [16.1K]

Answer:

The true weight of the aluminium is $m_{alu} =$ 4.5021 kg

Explanation:

Given data

$m_{app}$ = 4.5 kg

$\rho_{air}$ = 1.29 $\frac{kg}{m^{3} }$

$\rho_{al}$ = 2.7× $10^{3} \frac{kg}{m^{3} }$

The true mass of the aluminium is given by

$m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }$

Put all the values in above equation we get

$m_{alu} = \frac{(2700)(4.5)}{2700-1.29}$

$m_{alu} =$ 4.5021 kg

Therefore the true weight of the aluminium is $m_{alu} =$ 4.5021 kg

6 0

3 years ago

An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency o

kodGreya [7K]

Answer:

Part A

The efficiency of the engine is 10%

Part B

The change in internal energy is 300 J

Part C

The change in volume is 1 m³ which is one cubic meter

Explanation:

Part A

Efficiency is defined as the ratio of energy output to energy input;

The given parameters are;

The distance the car is moved, d = 10 meters

The force which moves the car, F = 100 N

The amount of energy produced by the engine = 1,000 J

Therefore, we have;

The energy output to the car = The work done on the car = Force applied to the car, F × The distance the car moves, d;

∴ The energy output to the car by the engine = F × d = 100 N × 10 m = 1,000 J

The energy input from the engine = The energy produced by the engine = 10,000 J

The efficiency of the engine = (The energy output)/(The energy input)= 1,000J/10,00J = 0.1

The efficiency in percentage = 0.1 × 100 = 10 %

The efficiency of the engine = 10%

Part B

The amount of heat added to the substance, ΔQ = 1,000J

The amount of work the substance does on the atmosphere, W = 700 J

The change in internal energy, ΔU is given as follows;

ΔQ = W + ΔU

∴ ΔU = ΔQ - W

For the substance, we have;

The change in internal energy, ΔU = 1,000 J - 700 J = 300 J

Part C

The work done by the piston, W = 1,000 J

The pressure, in the piston, P = 1,000 Pa = constant

The work done by the piston in a constant pressure process, W = P × ΔV

Where;

W = The work done

P = The constant pressure applied

ΔV = The change in volume = V₂ - V₁

V₂ = The final volume

V₁ = The initial volume

∴The change in volume ΔV = W/P = 1,000 J/(1,000Pa) = 1 m³

The change in volume ΔV = 1 m³

3 0

3 years ago