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Masja [62]
3 years ago
7

A swimmer moves through the water at 2.5 meters per second and experiences a drag force of 240 N. How much power is she generati

ng?
Physics
1 answer:
mote1985 [20]3 years ago
6 0
-- There's a force of 240N pushing her backwards.

-- She's maintaining a steady speed (of 2.5 m/s) .

-- In order to maintain a steady speed (no acceleration),
the forces on her must be balanced.  So she's maintaining
a steady force of 240N forward.

-- Every time she moves 1 m forward, she does work of
(force) x (distance) = 240 joules.

-- She moves 2.5 meters forward every second.
So she's doing (240 x 2.5) = 600 joules of work every second.

-- 600 joules per second = 600 watts .    
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When you voice the vowel sound in "hat," you narrow the opening where your throat opens into the cavity of your mouth so that yo
xz_007 [3.2K]

Answer: 0.1 m, 0.0583 m

Explanation:

We are given that:

Frequency for throat= 800 Hz

Frequency for mouth= 1500 Hz

Sound speed= 350 m/s

We have to find the corresponding lengths.

We have

f= \frac{v}{4L}

or L=\frac{v}{4f}

For the throat= L= \frac{350}{4*800\\} = 0.1 m

For the mouth= L= \frac{350}{4*1500} = 0.0583 m

3 0
3 years ago
How much total heat transfer is necessary to lower the temperature of 0.175 kg of steam from 125.5 °C to −19.5 °C, including the
GaryK [48]

Answer:

Explanation:

The heat required to change the temperature of  steam from 125.5  °C to 100 °C is:

Q_1 = ms_{steam} (125.5^0C - 100^0C) \\ \\ Q_1 = 0.175 \ kg ( 1520 \ J/kg.K ) (25.5^0 \ C) \\ \\ Q_1 = 6783 \ J

The heat required to change the steam at 100°C to water at 100°C is;

Q_2 = mL_v \\ \\ Q_2 = (0.175 \ kg) (2.25*10^6 \ J/kg ) \\ \\ Q_2 = 393750 \ J

The heat required to change the temperature from 100°C to 0°C is

Q_3 = ms_{water} (100^) \ C) \\ \\ Q_3 = (0.175\ kg)(4186 \ J/kg.K) (100 ^0c ) \\ \\ Q_3 = 73255 \ J

The heat required to change the water at 0°C to ice at 0°C  is:

Q_4 = mL_f \\ \\ Q_4 = (0.175 \ kg)(3.34*10^5 \ J/kg) \\\\ Q_4 = 58450 \ J

The heat required to change the temperature of ice from 0°Cto -19.5°C is:

Q_5 = ms _{ice} (100^0 C) \\ \\ Q_5 = (0.175 \ kg)(2090 \ J/kg.K)(19.5^0C)  \\ \\ Q_5 = 7132.125 \ J

The total heat required to change the steam into ice is:

Q = Q_1 + Q_2 + Q_3 + Q_4 +Q_5 \\ \\Q = (6788+393750+73255+58450+7132.125)J \\ \\ Q = 539325.125 \ J \\ \\ Q = 5.39*10^5 \ J

b)

The time taken to convert steam from 125 °C to 100°C is:

t_1 = \frac{Q_1}{P} = \frac{6738 \ J}{835 \ W}  = 8.12 \ s

The time taken to convert steam at  100°C to water at  100°C is:

t_2 = \frac{Q_2}{P} =\frac{393750}{834} =471.56 \ s

The time taken to convert water to 100° C to 0° C is:

t_3 = \frac{Q_3}{P} =\frac{73255}{834} = 87.73 \ s

The time taken to convert water at 0° to ice at 0° C is :

t_4 = \frac{Q_4}{P} =\frac{58450}{834} = 70.08  \ s

The time taken to convert ice from 0° C to -19.5° C is:

t_5 = \frac{Q_5}{P} =\frac{7132.125}{834} = 8.55  \ s

5 0
3 years ago
Which equation best describes the law of conservation of momentum?
NeX [460]
The law of conservation of momentum<span> states that for two objects colliding in an isolated system, the total </span>momentum<span> before and after the collision is equal. Momentum should be conserved. Hope this answers the question. Have a nice day.</span>
8 0
3 years ago
Read 2 more answers
The toy car is about 3 inches long is an example of a ?
balandron [24]

Answer:

a quantitative observation because it includes numerical data

8 0
2 years ago
A frame hanging on a wall is held by two cables. The tension in each cable is 30 N, and the cables make an angle of 45° with the
Dimas [21]

Answer: option D) 42.4 N

The weight of the frame is balanced by the vertical component of tension.

W = T sin θ + T sin θ = 2 T sin θ

The tension in each cable is T = 30 N

Angle made by the cables with the horizontal, θ = 45°

⇒ W = 2×30 N × sin 45° = 2 × 30 N × 0.707 = 42.4 N

Hence, the weight of the frame is 42.4 N. Correct option is D.


6 0
3 years ago
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