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3 years ago

You toss a marble straight up into the air with a speed of 1.3 m/s. How long does it take the marble to reach its highest point?

2 answers:

8 0

The acceleration of gravity is 9.8 m/s². That means that the speed of the marble is always 9.8 m/s greater downward than it was 1 second earlier.

You want to know when the marble is going to stop climbing. Well, that
will be when its upward velocity has dwindled to zero. That in turn is the
time when its upward velocity is 1.3 m/s less than it was when you tossed it.

(1.3 m/s) / (9.8 m/s²) = 0.133 second (rounded)

FromTheMoon [43]3 years ago

3 0

D
Reason:-
applying the formula
vf=vi+gt
1.3/9.8=t
0.13=t(Ans)

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andrezito [222]

Answer:

F = 0.00156[N]

Explanation:

We can solve this problem by using Newton's proposed universal gravitation law.

$F=G*\frac{m_{1} *m_{2} }{r^{2} } \\$

Where:

F = gravitational force between the moon and Ellen; units [Newtos] or [N]

G = universal gravitational constant = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

m1= Ellen's mass [kg]

m2= Moon's mass [kg]

r = distance from the moon to the earth [meters] or [m].

Data:

G = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

m1 = 47 [kg]

m2 = 7.35 * 10^22 [kg]

r = 3.84 * 10^8 [m]

$F=6.67*10^{-11} * \frac{47*7.35*10^{22} }{(3.84*10^8)^{2} }\\ F= 0.00156 [N]$

This force is very small compare with the force exerted by the earth to Ellen's body. That is the reason that her body does not float away.

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3 years ago

A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac

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Answer:

The force on one side of the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

$y=\dfrac{\Delta y}{\sin60}$

$y=\dfrac{2\Delta y}{\sqrt{3}}$

The area of strip is

$dA=\dfrac{9\times2\Delta y}{\sqrt{3}}$

We need to calculate the force on one side of the plate

Using formula of pressure

$P=\dfrac{dF}{dA}$

$dF=P\times dA$

On integrating

$\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}$

$F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}$

$F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})$

$F=3093529.3\ N$

Hence, The force on one side of the plate is 3093529.3 N.

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