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3 years ago

1) A common surface feature on the terrestrial planets is

1 answer:

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<span>1. A common surface features on the terrestrial planets is ___________.

Answer: A. Mountains and B. Volcanoes

2. Which scientist determined the planet's orbital shapes and sizes by triangulation?

Answer: C. Kepler

3. Scientists used the Keck telescope to learn more about the ______ located at the center of the Milky Way galaxy

Answer: A. star cluster

4. When the solar system formed, planetesimals collided and merged to form

Answer: A. Protoplanets

5) How is the heliocentric model of the solar system different than the geocentric model?

Answer: D. The heliocentric model has planets going around the sun; the geocentric model has planets going around Earth.</span>

You might be interested in

The charges that are free to move in a metallic conducting wire and that are responsible for the flow of electric current are- a

SVEN [57.7K]

Answer:

D) The negatively charged electrons

Electricity passes through metallic conductors as a flow of negatively charged electrons. The electrons are free to move from one atom to another. We call them a sea of delocalised electrons. Current was originally defined as the flow of charges from positive to negative. Please give me the brainliest answer?

:) Hoped this helped!!! Have a good day!!! <3

3 0

2 years ago

Write a short note on escape velocity.

andrezito [222]

Answer:

physics, escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body. The escape velocity from Earth is about 11.186 km/s (6.951 mi/s; 40,270 km/h; 25,020 mph)[1] at the surface. ... [2] Speeds higher than escape velocity have a positive speed at infinity.

6 0

2 years ago

Read 2 more answers

An auditorium measures 40.0 m 3 20.0 m 3 12.0 m. The density of air is 1.20 kg/m3. What are (a) the volume of the room in cubic

Burka [1]

We use the formula,

m = V\rho

Here, m is the mass, V is the volume and $\rho$ density

Also

$V = l w h$

Here l is length, w is width and h is height.

(a) The volume of the room,

$V=40.0 \ m \times 3 20.0 \ m \times 3 12.0\ m = 3.99 \times 10^{6} m^3$

The volume of the room in cubic feet,

$V = 3.99 \times 10^{6} m^3 \times(\frac{3.281 \ ft}{m} )^3 = 4.3 \times 10^7 \ ft^3$

(b) Now the mass of the air in room,

$m= (3.99 \times 10^{6} \ m^3) (1.20 \ kg/m^3) = 4.8 \times 10^6 kg$ .

Therefore, the weight of the air in room,

$W = mg= 4.8 \times 10^6 kg \times 9.8 m/s^2 = 46.9 \times 10^6 \ N \\\\ W = 4.69 \times 10^7 \ N$ .

The weight of air in the room in pounds,

$W = 4.69 \times 10^7 \ N ( \frac{1 \ lb}{4.448 \ N} ) = 1.1 \times 10^7 \ lb$

3 0

2 years ago

An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char

ArbitrLikvidat [17]

Answer:

Part a)

$W = 1.58 \times 10^{-20} J$

Part b)

$v = 1.86 \times 10^5 m/s$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = 4.00 \times 10^{-12} C/m^2$

now the electric field is given as

$E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}$

$E = 0.225 N/C$

Now acceleration of an electron due to this electric field is given as

$a = \frac{eE}{m}$

$a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}$

$a = 3.97 \times 10^{10}$

Now work done on the electron due to this electric field

$W = F.d$

$d = 0.470 - 0.03$

$d = 0.44 m$

So work done is given as

$W = (ma)(0.44)$

$W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)$

$W = 1.58 \times 10^{-20} J$

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

$W = \frac{1}{2}mv^2 - 0$

$1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2$

$v = 1.86 \times 10^5 m/s$

7 0

2 years ago

A 100-watt lightbulb uses 1 kilowatt-hour of electrical

Harlamova29_29 [7]

Answer:

Energy = 0.25 kilowatt-hour

Explanation:

Given the following data;

Power = 25 Watts

Time = 10 hours

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

$Power = \frac {Energy}{time}$

To find the energy consumed;

Energy = power * time

Substituting into the formula, we have;

Energy = 25 * 10

Energy = 250 Watt-hour

To convert to kilowatt-hour, we would divide by 1000;

Energy = 250/1000

Energy = 0.25 kilowatt-hour

3 0

2 years ago