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3 years ago

10

A cheerleader jumps off from the shoulders of two of her teammates. They all fell at the same rate, as a result. This type of co

llision is...?

1 answer:

melamori03 [73]3 years ago

5 0

Answer:

elastic collision

Explanation:

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1. A rock of granite has a mass of 50 kg. if it’s weight in water

igor_vitrenko [27]

Answer:

The first part of the question is asking about BUOYANT FORCE or UPTHRUST.

Upthrust =TRUE WEIGHT-APPARENT WEIGHT

TRUE WEIGHT=mg

TRUE weight=50kg×10m/s²

=500N

upthrust=500N-380N

FB=120N

volume of the rock=mass/density.

since the granite is completely submerged, the volume of the displaced liquid will be equal to the volume of the body.

upthrust=Vdg

120N=V×1000kg/m³×10m/s²

120N=V×10000kg/m²s²

120/10000=V

v=0.012m³

please mark brainliest, hope it helped

6 0

3 years ago

What is the unit for force?

Morgarella [4.7K]

Answer:

N / NEWTONS

Explanation:

Named after Isaac Newton, the man who discovered gravity

3 0

3 years ago

Read 2 more answers

At what water temperature will additional heat energy need to be added before the temperature will change again?

Katyanochek1 [597]

That would be 0 degrees Celsius aka the melting point of water.... If you look at the diagram I attached you notice that at 0 degrees Celsius it is flat, this is because much heat is needed at this point for water to rise to 1 degree... It is the same for the boiling point (100)

Hope this helps! (If correct, please rank as brainliest answer) :)

7 0

3 years ago

A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o

Crank

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

$\delta = \frac{PL}{AE}$

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

$P = 68*10^3 N$

E = 200GPa

$A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2$

Replacing we have,

$\delta = \frac{PL}{AE}$

$\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}$

$\delta = 0.001932m$

$\delta = 1.93mm$

Therefore the change in length is 1.93mm

7 0

3 years ago

Please help ASAP! I will give the first answer brainliest!

Natalija [7]

Answer:

Sorry for being late. Hope this helps

Explanation:

3 0

3 years ago