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Ainat [17]
3 years ago
9

Each mass in the figure is 3 kg. Find the magnitude and direction of the net gravitational force on mass A due to the other mass

es.
A. 2.45 × 10–7 N toward B
B. 3.75 × 10–7 N toward C
C. 2.00 × 10–7 N toward D
D. 1.15 × 10–7 N toward D

Physics
2 answers:
amm18123 years ago
4 0

Answer:

THE ANSER IS B

Explanation:

svet-max [94.6K]3 years ago
4 0
The answer is B because it’s the way it is
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A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water
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Answer:

(a) 1.6m

(b) 6.86m

(c) 2.86s

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

x(t)=x_{0}+v_{0}t

Where x(t) is the horizontal distance at a time t, x_{0} is the initial horizontal position which in this case will be zero: x_{0}=0. And v_{0} is the initial speed: v_{0}=2m/s

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: t=0.8s so the horizontal distance is:

x(0.8)=(2m/s)(0.8s)1.6m

<u>The answer for (a) is 1.6m</u>

Now, to solve (b) we need the equation for vertical distance:

y(t)=y_{0}-\frac{1}{2}gt^2

Where y(t) is the vertical distance at a time t, y_{0} is the initial vertical distance: y_{0}=10m And g is the gravitational acceleration: g=9.81m/s^2 }

(b) At what vertical distance above the surface of the water is the diver just then?

The time is the same as in the last question: t=0.8s

y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2

y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)

y(0.8)=10m-\frac{1}{2}(6.2784m)

y(0.8)=10m-(3.1392)

y(0.8)=6.86m

<u>the answer to (b) is 6.86m</u>

(c) At what horizontal distance from the edge does the diver strike the water?

To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0

Using y(t)=y_{0}-\frac{1}{2}gt^2

if y(t)=0

0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t

and since g=9.81m/s^2

t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s

At t=1.43s the car hits the water, so the horizontal distance at that time is:

x(t)=x_{0}+v_{0}t

x(1.43)=(2m/s)(1.43s)=2.86m

<u>the answer to (c) is 2.86s</u>

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