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3 years ago

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Select the correct answer.

1 answer:

Dovator [93]3 years ago

3 0

Answer:

D. Mars is a long distance away

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Name at least four factors involved in performing a good throw.

atroni [7]

Answer:

You need to have a good stance, release, power, speed, body, movement, and aim. You should be able to throw an effective pass or throw by using your body and having the correct amount of balance and form.

Explanation:

7 0

3 years ago

Please Help No Links

ale4655 [162]

Answer:

the moment of inertia

Explanation:

5 0

3 years ago

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

VladimirAG [237]

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>

6 0

3 years ago

Three batteries are connected in series so that the total voltage is 54 volts. The voltage of the first battery is twice the vol

Alex_Xolod [135]

Answer:

$v_1 = 12 volts$

$v_2 = 6 volts$

$v_3 = 36 volts$

Explanation:

As we know that all the batteries are in series

so the net voltage of all three batteries is given as

$V = v_1 + v_2 + v_3$

now we know that

$v_1 = 2v_2$

$v_1 = \frac{1}{3}v_3$

now plug in all the values in it

$54 = v_1 + \frac{v_1}{2} + 3v_1$

$54 = 4.5 v_1$

$v_1 = 12 volts$

now we have

$v_2 = 6 volts$

$v_3 = 36 volts$

3 0

3 years ago

Calculate the electrical energy expended in a device across which the circuit voltage drops by 50 volts in moving a charge of 6

solmaris [256]

1 volt = 1 joule per coulomb 50 volts = 50 joules per coulomb 50 joules/coulomb times 6 coulombs = 300 joules

5 0

4 years ago

Read 2 more answers