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Sunny_sXe [5.5K]

4 years ago

10

Dense fluids exert___________buoyant force then less dense fluids because they have more mass within a certain volume than a les

s dense fluid.

1 answer:

sergij07 [2.7K]4 years ago

6 0

Answer:

The answer is 'more' as more mass can exert more pressure

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The gravitational force between two objects that are 2.1 times 10^-1 m apart is 3.2 times 10^-6 N. If the mass of one object is

Shtirlitz [24]

Answer:

Mass of the other object is 38.45 kg.

Explanation:

Given:

The gravitational force between two objects is, $F=3.2\times 10^{-6}$ N

Mass of one object is, $m_{1}=55\ kg$

Distance between the objects is, $r=2.1\times 10^{-1}\ m$

Gravitational constant is, $G =6.674\times 10^{-11}\ m^3 kg^{-1}s^{-2}$

Let the mass of the other object be $m_{2}\ kg$

Gravitational force is given as:

$F=\frac{Gm_1m_2}{r^2}$

Plug in the given values and solve for $m_2$ . This gives,

$3.2\times 10^{-6}=\frac{6.674\times 10^{-11}\times 55\times m_2}{(2.1\times 10^{-1})^2}\\3.2\times 10^{-6}=\frac{6.674\times 10^{-11}\times 55\times m_2}{0.0441}\\3.2\times 10^{-6}=8.3236\times 10^{-8}m_2\\m_2=\frac{3.2\times 10^{-6}}{8.3236\times 10^{-8}}= 38.45\ kg$

Therefore, the mass of the other object is 38.45 kg.

3 0

4 years ago

How much heat is needed to melt a 3 kg block of cheddar cheese initially at room tmperature of 24 degrees C? (assume the melting

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Cheese I'm so sorry for not having an answer I failed you

8 0

4 years ago

A 45.0-N force pushes a cart 12.5 meters down a hallway. What is the work done on the cart?

makvit [3.9K]

Answer:

562.5J

Explanation:

The following were obtained from the question:

F = 45N

d = 12.5m

w =?

The work done can be achieved by using

w = F x d

w = 45 x 12.5

w = 562.5J

3 0

3 years ago

Two rams run toward each other. One ram has a mass of 49 kg and runs west

olga nikolaevna [1]

Answer: (d)

Explanation:

Given

Mass of the first ram $m_1=49\ kg$

The velocity of this ram is $v_1=-7\ m/s$

Mass of the second ram $m_2=52\ kg$

The velocity of this ram $v_2=9\ m/s$

They combined after the collision

Conserving the momentum

$\Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow 49\times (-7)+52\times (9)=(52+49)v\\\Rightarrow v=\dfrac{125}{101}\ m/s \quad[\text{east}]$

Momentum after the collision will be

$\Rightarrow 101\times \dfrac{125}{101}=125\ kg-m/s\ \text{East}$

Therefore, option (d) is correct

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3 years ago

) or<br> The best places to take a pulse are at the radial artery (<br> the carotid artery (

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Answer:

on edge, it is all of them:

carotid

brachial

radial

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dorsalis pedis

Explanation:

7 0

3 years ago