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never [62]
3 years ago
10

It will be easier to lift a load in wheel barrow if the load is movedtowards the wheel.why​

Physics
1 answer:
Trava [24]3 years ago
5 0

Answer:

the moment

Explanation:

the shorter the distance from the force, the lesser the torque acting on it

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Wire Materials: The most common material for electrical wire is copper and aluminum, these are not the best conductors however they are abundant and low cost. Gold is also used in applications because it is corrosion resistant.
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what happens to the water after it rains? a. precipitation b. runoff c. condensation user: what do organisms from the ocean use
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The rain gets evaporated in to water vapor and is returned to the clouds where they go through condensation and then they poud down as rain or A.K.A,  Precipitation.
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A golf ball traveling 3m/s to the right collides in a head-on collision with a stationary bowling ball in a friction-free enviro
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Answer: equal to 3 m/s

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So speed of golf ball will be same in order to Satisfy

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3 0
4 years ago
From the top of the leaning tower of Pisa, a steel ball is thrown vertically downwards with a speed of 3.00 m/s. if the height o
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66,7 seconds

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3 0
3 years ago
The radioactive element radium (Ra) decays by a process known asalpha decay,in which the nucleus emits a helium nucleus. (Factoi
tamaranim1 [39]

<u>Answer:</u> The energy released in the decay process is 4.6800\times 10^{11}J

<u>Explanation:</u>

The equation for the alpha decay of Ra-226 follows:

_{88}^{226}\textrm{Ra}\rightarrow _{2}^{4}\textrm{He}+_{86}^{222}\textrm{Rn}

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

\Delta m=(m_{Rn}+m_{He})-(m_{Ra})

We know that:

m_{Rn}=222.0176u\\m_{Ra}=226.0254u\\m_{He}=4.0026u

Putting values in above equation, we get:

\Delta m=(222.0176+4.0026)-(226.0254)=-0.0052g=-5.2\times 10^{-6}kg

(Conversion factor: 1 kg = 1000 g )

To calculate the energy released, we use Einstein equation, which is:

E=\Delta mc^2

E=(-5.2\times 10^{-6}kg)\times (3\times 10^8m/s)^2

E=-4.6800\times 10^{11}J

Hence, the energy released in the decay process is 4.6800\times 10^{11}J

8 0
3 years ago
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