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3 years ago

What is the force that gravity puts upon a 25kg person

Physics

1 answer:

yan [13]3 years ago

3 0

Answer:

245.25 Newtons

Explanation:

the gravitational force of the earth is 9.81 N/Kg

And so to work this out you would multiply 25 by 9.81 to give you 245.25

$25*9.81=245.25N$

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An electron is pushed into an electric field where it acquires a 1-v electric potential. if two electrons are pushed the same di

liq [111]

Given here that through the same region 2 electrons are pushed by same distance

As we know that electric potential can be relate with electric field by following relation

$\delta V = E . d$

Since the two electrons are pushed through same electric field by same distance

So the product of E and d will be same

So the electric potential will be same as initial

so V = 1 volt for two electrons

6 0

4 years ago

A satellite is in a circular orbit around the Earth at an altitude of 2.80 3 106 m. Find (a) the period of the orbit, (b) the sp

Katen [24]

Explanation:

Given that,

Altitude $h= 2.803\times10^{6}\ m$

We need to calculate the radius

$r=R+h$

Where, R = radius of the earth

h = radius of altitude

Put the value into the formula

$r=(6.38\times10^{6}+2.803\times10^{6})$

$r=9.18\times10^{6}\ m$

(a). We need to calculate the period of the orbit,

Using formula of period

$T^2=\dfrac{4\pi^2}{GM}r^3$

$T^2=\dfrac{4\pi^2}{6.67\times10^{-11}\times5.98\times10^{24}}\times(9.18\times10^{6})^3$

$T^2=76570372.9509\ sec$

$T=8750.45\ sec$

(b). We need to calculate the speed of the satellite

Using formula of speed

$v^2=\dfrac{GM}{r}$

Put the value into the formula

$v^2=\dfrac{6.67\times10^{-11}\times5.98\times10^{24}}{9.18\times10^{6}}$

$v^2=43449455.3377\ m/s$

$v=6.59\times10^{3}\ m/s$

(c). We need to calculate the acceleration of the satellite

Using formula of acceleration

$a_{c}=\dfrac{v^2}{r}$

Put the value into the formula

$a_{c}=\dfrac{(6.59\times10^{3})^2}{9.18\times10^{6}}$

$a_{c}=4.73\ m/s^2$

Hence, This is the required solution.

6 0

4 years ago

Explain retrograde motion and why it confused early astronomers as they viewed the motion of objects in the sky.

Masteriza [31]

Answer:

Explanation:

Retrograde motion in astronomy is, in general, orbital or rotational motion of an object in the direction opposite the rotation of its primary, that is, the central object (right figure). It may also describe other motions such as precession or nutation of an object's rotational axis. Prograde or direct motion is more normal motion in the same direction as the primary rotates. However, "retrograde" and "prograde" can also refer to an object other than the primary if so described. The direction of rotation is determined by an inertial frame of reference, such as distant fixed stars.

3 0

3 years ago

Read 2 more answers

A mercury manometer is connected on one side to a bulb containing argon, while the other end is open to atmospheric pressure, wh

belka [17]

Answer:

740, mm of Hg

Explanation:

The pressure of argon , in mm of Hg = difference in the level of mercury on either side of the manometer.

mercury column in the open end of the manometer is 22 mm below that in the side connected to the argon and 762 mmHg, end is open to atmospheric pressure.

therefore, The pressure of argon , in mm of Hg =762 -22 = 740, mm of Hg

4 0

3 years ago

Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$