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3 years ago

13

Suppose a person sits on a skateboard with her feet up and throws a ball. Explain why she will move as a result of throwing the

ball.

1 answer:

Snowcat [4.5K]3 years ago

8 0

I’m guessing is because she uses force to throw the ball, allowing the energy to move the person.

sorry if it’s not 100% correct

You might be interested in

You move a 4 N object 10 meters. Find work

Trava [24]

Answer: 40 J

Explanation: Work is equal to the product of force and distance.

W = Fd

= 4N ( 10 m)

= 40 J

8 0

3 years ago

Identify and sketch all the external forces acting on the chair. Because the chair can be represented as a point particle of mas

Monica [59]

Answer:

y axis normal (N) and the weight (W)

x axis pplied force (F) and friction force (fr)

Explanation:

If we have a chair on a horizontal surface, the normal (N) and the weight (W) of the body act on the vertical axis.

On the x axis, the applied force (F) acts in the direction of movement and the friction force (fr) in the opposite direction of movement.

In this exercise we assume that the body tends to move to the right, all the forces can be seen in the adjoint

5 0

2 years ago

Why are the Soviets ahead in the race to capture the V-2 rocket and Werner von Braun?

Dahasolnce [82]

The second world war, and its war weapons, such as the v-2 rockets, had a great impact on the world until today, to answer this question we need that...

<h3>V-2 factory </h3>

On April 11, 1945, US troops took the town of Bleicherode, in the Kohnstein region, where the V-2 factory was located. From there about 100 complete V-2s and thousands of parts and equipment were "captured" as war loot and transferred to the United States, where they formed the basis for practical studies of the missile defense program.

With this information, we can say that because it was a base discovered by US troops, none of the alternatives is correct, as it was not the Soviets who discovered it, and that the base was also located in the central part of Germany.

<u>The </u><u>v-2 factory</u><u> was located in </u><u>Kohnstein, central germany,</u><u> by this claim, </u><u>none of the alternatives is correct.</u>

Learn more about second world war brainly.com/question/7013432

5 0

2 years ago

An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst

sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

$T_1 =$ 500°C

$T_2$ = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

$\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5$

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to $T_{avg}$ = 535.5K $\approx 550K$

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

$P_r = 0.63$

A)

Number for the first strips is equal to

$R_e_x = \frac{u_o.L}{v}$

$R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4$

Calculating heat transfer coefficient from the first strip

$h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3$

$h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2$

The rate of convection heat transfer from the first strip is

$q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W$

The rate of convection heat transfer from the fifth trip is equal to

$q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2$

Calculating $h_o_-_4$

$h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2$

The rate of convection heat transfer from the tenth strip is

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2$

Calculating

$h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W$

The rate of convection heat transfer from 25th strip is equal to

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)$

Calculating $h_o_-_2_5$

$h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2$

Calculating $h_o_-_2_4$

$h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W$

6 0

3 years ago

9. A wave on Beaver Dam Lake passes by two docks that are 40.0 m apart.

Kobotan [32]

Answers:

a) 10 m

b) time=1.6 s, frquency=0.625 Hz

c) 6.25 m/s

Explanation:

a) If there is a crest at each dock and another three crests between the two docks, and the wavelength $\lambda$ is the distance between to crests; this means we have $4\lambda$ in $40 m$ :

$40 m=4\lambda$

Clearing $\lambda$ :

$\lambda=\frac{40 m}{4}$

$\lambda=10 m$

b) This part can be solved by a Rule of Three:

If 10 waves ---- 16 s

1 wave ----- $T$

Then:

$T=\frac{(1 wave)(16 s)}{10 waves}$

$T=1.6 s$ This is the period of the wave

On the other hand, the frequency $f$ of the wave has an inverse relation with its period $T$ :

$f=\frac{1}{T}$

$f=\frac{1}{1.6 s}$

$f=0.625 Hz$ This is the frequency of the wave

c) The speed $v$ of a wave is given by the following equation:

$v=\frac{\lambda}{T}$

$v=\frac{10 m}{1.6 s}$

Finally:

$v=6.25 m/s$

4 0

3 years ago