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3 years ago

PLEASE HELP!! I'll mark brainliest for correct answer.

Physics

1 answer:

Tju [1.3M]3 years ago

6 0

1. Car A
2. Car C
3. Car A
4. 1500 J
5. 750 J
6. 750 J
7. Car A was had a height higher relative to the ground compared to all the other cars.
8. mgh
9. B

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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

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

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:

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

The answer for (a) is 1.6m

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: $t=0.8s$

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

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

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

$y(0.8)=10m-(3.1392)$

$y(0.8)=6.86m$

the answer to (b) is 6.86m

(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:

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

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

the answer to (c) is 2.86s

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3 years ago

What is a charge-coupled device (CCD), and how is it used in astronomy?

Sedaia [141]

Answer:

Charge-coupled device (CCD) is a device that receives and transfers an electrical charge to the next region

Explanation:

Charge-coupled device (CCD) is a device that receives and transfers an electrical charge to the next region where it can be modified like changing it to a electronic value.

In astronomy, high-powered telescopes can be used with CCD device image sensor cameras. The imaging system can concentrate for a number of hours on one place in space once the Earth's rotation synchronizes with the telescope.

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4 years ago

Calculate the entropy change that occurs when 1.0kg of water at 20.00 C is mixed with 2.0kg of water at 80.00 C

SOVA2 [1]

Answer:

The change in entropy ΔS = 0.0011 kJ/(kg·K)

Explanation:

The given information are;

The mass of water at 20.0°C = 1.0 kg

The mass of water at 80.0°C = 2.0 kg

The heat content per kg of each of the mass of water is given as follows;

The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg

The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg

Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg

The heat energy of the mixture =

1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)

∴ T = 60°C

The heat content, of the water at 60° = 251.154 kJ/kg

Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462

The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).

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3 years ago

From the above diagram, which phase of the Moon will result in the greatest difference between high and low tide? A. New Moon B.

aliya0001 [1]

New moon, because the New Moon (A) because the sun and the moon work together most when it is there on the tides

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3 years ago

Read 2 more answers

This is the change in kinetic energy of a system in which a 16 kg object moving at 25 m/s slows to a velocity of 20 m/s

Dennis_Churaev [7]

The kinetic energy of an object is given by

KE = 0.5mv²

where m is the mass and v is the velocity.

To calculate the change in kinetic energy...

Initial KE:

KEi = 0.5mVi²

where Vi is the initial velocity.

Final KE:

KEf = 0.5mVf²

where Vf is the final velocity.

ΔKE = KEf - KEi

ΔKE = 0.5mVi² - 0.5mVf²

ΔKE = 0.5m(Vf²-Vi²)

Given values:

m = 16kg

Vi = 25m/s

Vf = 20m/s

Plug in the given values and solve for ΔKE:

ΔKE = 0.5×16×(20²-25²)

ΔKE = -1800J

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3 years ago