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3 years ago

The movement of bones around a fixed angle without lateral displacement is called

1 answer:

3 0

The answer to the question being asked on what do you called when the movement of the bones around a fixed angle without lateral displacement is ROTATION. Rotation movement allows the bone to move in a single long axis without being displaced.

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If earth were a ping pong what size ball would jupiter be

Mariulka [41]

Answer:

a large beach ball

Explanation:

6 0

2 years ago

The Sun is about 150 million km from earth. how long does it take light from the sun to reach earth? (Speed of light is 3x10^8m/

madam [21]

Time =      (distance) / (speed)

Time = (150 x 10⁹ m) / (3 x 10⁸ m/s) =

   50 x 10¹ sec =

       <em>500 sec</em> = 8 min 20 sec

3 0

3 years ago

. A weather balloon is inflated to a volume of 2.2 x 103 L with 37.4 g of helium. What is the density of helium in grams per lit

Margarita [4]

That is the mst best eway to find its solution.
37.4/2.2*10^3 = 0.017 gm/liter or 1.7*10^-2
so we conclude that option b is sorrect

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3 years ago

Hydrogen-2 and Hydrogen-3 fuse to form Helium-4 and a neutron. How much energy is released in this nuclear reaction?

IgorC [24]

1 jul is released after

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3 years ago

A nonconducting container filled with 25 kg of water at 23°C is fitted with a stirrer, which is made to turn by gravity acting o

Paul [167]

Explanation:

Given that,

Weight of water = 25 kg

Temperature = 23°C

Weight of mass = 32 kg

Distance = 5 m

(a). We need to calculate the amount of work done on the water

Using formula of work done

$W=mgh$

$W=32\times9.8\times5$

$W=1568\ J$

The amount of work done on the water is 1568 J.

(b). We need to calculate the internal-energy change of the water

Using formula of internal energy

The change in internal energy of the water equal to the amount of the work done on the water.

$\Delta U=W$

$\Delta U=1568\ J$

The change in internal energy is 1568 J.

(c). We need to calculate the final temperature of the water

Using formula of the change internal energy

$\Delta U=mc_{p}\Delta T$

$\Delta U=mc_{p}(T_{2}-T_{1})$

$T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}$

$T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}$

$T_{2}=23.01^{\circ}\ C$

The final temperature of the water is 23.01°C.

(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.

The amount of heat is 1568 J.

Hence, This is the required solution.

6 0

3 years ago