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3 years ago

When a water wave runs into you at the beach, what causes you to get knocked down?

Physics

1 answer:

Dominik [7]3 years ago

8 0

The answer should be A

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Who is gonna win the super bowl patriots or Seahawks

Oksana_A [137]

The NE Patriots will. Just you wait and see. And when thay do, remember
that you saw the correct prediction right here on Brainly, posted by AL2006.

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4 years ago

Read 2 more answers

Tarzan swings back and forth on a

WARRIOR [948]

Answer:

Considering Tarzan as a point mass at point L and using the equation for a simple pendulum:

P = 2 pi (L / g)1/2

P^2 = 4 pi^2 L / g

L = g P^2 / (4 pi^2) = 9.8 m/s^2 * 7.27^2 s^2 / (4 pi^2) = 13.1 m

4 0

3 years ago

Two balls of different masses are dropped from a building, and they hit the ground at the same time.

Reil [10]

Answer:

speed and velocity, c would be my guess

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3 years ago

A rectangular plate has a length of (21.7 ± 0.2) cm and a width of (8.2 ± 0.1) cm. Calculate the area of the plate, including it

Grace [21]

Answer:

(177.94 ± 3.81) cm^2

Explanation:

l + Δl = 21.7 ± 0.2 cm

b + Δb = 8.2 ± 0.1 cm

Area, A = l x b = 21.7 x 8.2 = 177.94 cm^2

Now use error propagation

$\frac{\Delta A}{A}=\frac{\Delta l}{l}+\frac{\Delta b}{b}$

$\frac{\Delta A}{A}=\frac{0.2}{21.7}+\frac{0.1}{8.2}$

$\Delta A=177.94 \times \left ( 0.0092 + 0.0122 \right )=3.81$

So, the area with the error limits is written as

A + ΔA = (177.94 ± 3.81) cm^2

8 0

3 years ago

A Carnot engine operates between temperature levels of 600 K and 300 K. It drives a Carnot refrigerator, which provides cooling

KATRIN_1 [288]

Explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

$\frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}$ ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

$\frac{W}{Q_{C_{2}}} = \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}}$ ...... (2)

Now, equating both equations (1) and (2) as follows.

$Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}}$ = $Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}$

$\gamma = \frac{Q_{C_{2}}}{Q_{H_{1}}}$

= $\frac{T_{C_{2}}}{T_{H_{1}}} (\frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{2}} - T_{C_{2}}})$

= $\frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})$

= 2.5

Thus, we can conclude that the ratio of heat extracted by the refrigerator ("cooling load") to the heat delivered to the engine ("heating load") is 2.5.

4 0

3 years ago