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Verdich [7]
3 years ago
13

Planet Kling has half the radius and 2 times the mass of the Earth. What is the best estimate for the magnitude of the gravitati

onal field at the surface of planet Kling?
Physics
2 answers:
Vlad [161]3 years ago
6 0

Answer:80 m/s^2

Explanation:

alekssr [168]3 years ago
3 0

Answer:

80m/s²

Explanation:

First, use the equation below to find the strength of a planet/moon's satellite field and acceleration rate.

This is G × Mass of planet/moon ÷ r² (which is the distance or conversely radius squared)

Then you want to plg in all your values, which are going to be the gravitational constant, Kling's mass, and Kling's radius.

Remember that Earth's mass is equal to 5.97 × 10∧24 kg and its radius is exactly 6,731 km, so this will help you figure out your values for any other problems like this .

Also that the big ( G) you see is = 6.673×10∧-11.

Now for step two, which is the last thing you need to do. Once you've simplified the above equation, take the value you found, which should ≈ 7.85, and then multiply it by the force of gravity on Earth.

Once you multiply 7.85 by 9.8 m/s², the answer will be = 76.93 or ≈ 80m/s².

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lara [203]
I think the answer is B
6 0
3 years ago
Several springs are connected as illustrated below in (a). Knowing the individual springs stiffness k1 = 20 N/m, k2 = 30 N/m, k3
Hatshy [7]

Answer:

The equivalent stiffness of the string is 8.93 N/m.

Explanation:

Given that,

Spring stiffness is

k_{1}=20\ N/m

k_{2}=30\ N/m

k_{3}=15\ N/m

k_{4}=20\ N/m

k_{5}=35\ N/m

According to figure,

k_{2} and k_{3} is in series

We need to calculate the equivalent

Using formula for series

\dfrac{1}{k}=\dfrac{1}{k_{2}}+\dfrac{1}{k_{3}}

k=\dfrac{k_{2}k_{3}}{k_{2}+k_{3}}

Put the value into the formula

k=\dfrac{30\times15}{30+15}

k=10\ N/m

k and k_{4} is in parallel

We need to calculate the k'

Using formula for parallel

k'=k+k_{4}

Put the value into the formula

k'=10+20

k'=30\ N/m

k_{1},k' and k_{5} is in series

We need to calculate the equivalent stiffness of the spring

Using formula for series

k_{eq}=\dfrac{1}{k_{1}}+\dfrac{1}{k'}+\dfrac{1}{k_{5}}

Put the value into the formula

k_{eq}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}

k_{eq}=8.93\ N/m

Hence, The equivalent stiffness of the string is 8.93 N/m.

3 0
3 years ago
Which feature of an object affects its weight? Select three options.
Vadim26 [7]

Answer:mass of the object,how much force earth exerts on the object,and shape of the object

Explanation:

6 0
3 years ago
What building materials do you believe would work well to build a home in that area? Explain why you chose these materials.
Luden [163]
Depends on what the area is. If it’s a rural place, Wood is cheep & easy to build. If there’s a lot of corrosion, strong weather/hurricane, bricks.
8 0
4 years ago
A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the
11Alexandr11 [23.1K]

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\

<u>v_f = 10.85 m/s</u>

4 0
3 years ago
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