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mario62 [17]
4 years ago
13

When is your weight is equal to mg?​

Physics
2 answers:
Vinil7 [7]4 years ago
6 0

Answer:

The weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity, w = mg.

Sergio [31]4 years ago
3 0

Answer:

Weight is always equal to mg.

as we know the formula

Weight=mass(m)× acceleration due to gravity (g)

Your may may be when our weight is equal to Zero.

Our weight is zero:1) at the center of the earth because value of g is 0 there

2)in the space at null point

3)during weightlessness

i guess it may be your required answer.please have a look and reply

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2<br> 7.42 X 10<br> What is the scientific notation for this equation
galben [10]

Answer:

7 42 x 10^1

Explanation:

Because 10 x 1 =10

6 0
3 years ago
What force controls the isostatic adjustment of Earth’s crust?
Alina [70]
The force that control the isostatic adjustment of the earth's crust is THE FORCE OF GRAVITY.
The isostatic adjustment of the earth crust refers to continuous movement of the earth crust or its deformation as a result of the force of gravity which act on it. The earth crusts that are located at the upper and a the core of the earth usually set up a balance between them which is called isostatic equilibrium.
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3 years ago
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A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
3 years ago
A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration
JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on  and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m

So, the maximum height reached by the rocket is:

h=y_1+y_2\\h=160m+154.28m=314.28m

(c) In the first part we have:

v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s

And in the second part:

t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s

So,  the time it takes to reach the maximum height is:

t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}

t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s

So, the total time is:

t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s

7 0
4 years ago
The depth of a pond is 1.5m. Calculate the pressure caused by the water at the bottom of the pond ?​?
zepelin [54]

Answer:

Area=1.5(1.5)=2.25m^2

Force of gravity=10N

\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}

⟼Pressure=

Area

Force

\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}

⟼Pressure=

2.25

10

\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}

⟼Pressure=4.4Pa

5 0
3 years ago
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