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loris [4]
2 years ago
11

Is it okay to keep an albino gecko

Physics
1 answer:
arlik [135]2 years ago
5 0

Answer:

Yes

Explanation:

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nan moved 18m to the right and then another 22m to the right if the motion takes 20 seconds what is nans velocity
IrinaVladis [17]
Step 1: list known info

distance(change in position (Δx))= 18m+22m= 40m
time= 20 seconds

Step 2 :solve for velocity

velocity= Δx÷time
v= 40/20= 2m/s

Answer: the velocity is 2 meters per a second (m/s)


7 0
3 years ago
The tilt of the Earth's axis of rotation changes from approximately 22.5° to 24.5° over a period of 41,000 years. This causes gl
kompoz [17]
<span>b. less climatic variation between the summer and winter seasons in the middle and high latitudes

As the tilt becomes higher (approaches 24 degrees) there is greater variation between the summer and winter months, due to the fact that the tilt toward the sun in the summer and away from the sun in the winter are more pronounced. </span>
3 0
3 years ago
Read 2 more answers
Calculate the force between two objects that have masses of 70 Kilograms and 2,000 kilograms separated by a distance of 1 meter
Hunter-Best [27]

9.3 x 10⁻⁶N

Explanation:

Given parameters:

Mass 1 = 70kg

Mass 2 = 2000kg

distance = 1m

Unknown:

force between them =  

Solution:

The force between the two masses will be a gravitational force of attraction.

  F = \frac{G m_{1}m_{2}  }{r^{2} }

 G is universal gravitation constant = 6.67430×10−¹¹ N⋅m²/kg²

 r is the distance between the two masses

Substituting the parameters:

 F = \frac{6.67 x 10^{-11} x 70 x 2000}{1^{2} } = 9.3 x 10⁻⁶N

 Learn more:

Universal gravitation constant   brainly.com/question/1724648

#learnwithBrainly

5 0
3 years ago
Bacteria which do not live in extreme conditions are found in which kingdom:
mel-nik [20]
Part of the
archeabactria
8 0
3 years ago
State Pascal’s law. (2) b) The area of one end of a U-tube is 0.01 m2 and that of the other
Ray Of Light [21]

Answer:

Pascal Law's says that:

If the area of one end of a U-tube is A, and the area of the other end is A'. then if we apply a force F in the first end (the one of area A), the force experienced at the other end must be:

F' = F*(A'/A).

b) Now we can apply this to our particular case:

if the area of one end is 0.01m^2, and the area of the other end is 1m^2

Then we have:

A = 0.01m^2

A' = 1m^2

So, if now we apply a force F in the first end, the force experienced at the other end will be:

F' = F*(1m^2/0.01m^2) = F*100

This means that the force in the other end must be 100 times the force in the first end.

3 0
3 years ago
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