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3 years ago

An object of mass 82kg is accelerated upward at 3.2m/s/s. what force is required

Physics

1 answer:

vampirchik [111]3 years ago

6 0

F = m · a

In order to accelerate 82 kg upward at the rate of 3.2 m/s², a NET upward force of (82kg · 3.2m/s²) = 262.4 Newtons is required.

But if the object is on or near the surface of the Earth, then there's a downward force of (82kg · 9.8m/s²) = 803.6 N already acting on it because of gravity.

So you need to apply (803.6N + 262.4N) = <em>1,066 Newtons UPward</em>, in order to cancel its own weight and accelerate it upward at that rate.

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Refraction occurs when __________.

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3 years ago

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3 years ago

How much power does a hair dryer require to transform 198 000 J of energy in 15 mins?

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3 years ago

Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi

Andreas93 [3]

Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite $r_{1}= 8.20\times10^{7}$

Orbital radius of second satellite $r_{2}=7.00\times10^{7}\ m$

Mass of first satellite $m_{1}=53.0\ kg$

Mass of second satellite $m_{2}=54.0\ kg$

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

$v=\sqrt{\dfrac{GM}{r}}$

From this relation,

$v_{1}\propto\dfrac{1}{\sqrt{r}}$

Now, $\dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}$

$v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}$

Put the value into the formula

$v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}$

$v_{2}=5195.16\ m/s$

Hence, The orbital speed of this second satellite is 5195.16 m/s.

4 0

3 years ago

Umar has two copper pans, each containing 500cm3 of water. Pan A has a mass of 750g and pan B has a mass of 1.5kg. Which pan wil

Olin [163]

Answer:

heat required in pan B is more than pan A

Explanation:

Heat required to raise the temperature of the substance is given by the formula

$Q = ms\Delta T$

now we know that both pan contains same volume of water while the mass of pan is different

So here heat required to raise the temperature of water in Pan A is given as

$Q_1 = (m_w s_w + m_ps_p)\delta T$

$Q_1 = (0.5(4186) + 0.750(s))\Delta T$

Now similarly for other pan we have

$Q_2 = (m_w s_w + m_ps_p)\delta T$

$Q_2 = (0.5(4186) + 1.50(s))\Delta T$

So here by comparing the two equations we can say that heat required in pan B is more than pan A

3 0

3 years ago