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RUDIKE [14]
3 years ago
13

A skydiver has a mass of 110 kg. At what speed will she have a momentum

Physics
2 answers:
Alex17521 [72]3 years ago
7 0

Answer:

\boxed{B}

Explanation:

v=\frac{p}{m}

v=velocity\\p=momentum\\m=mass

v=?\\p=1000\\m=110

v=\frac{1000}{110}

v \approx 9.09

bogdanovich [222]3 years ago
6 0
Yea I’m pretty sure b
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A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete
NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}

From Rayleigh criteria

sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm

The focal length of the camera's lens is 300cm

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3 years ago
What is the great red spot? what is the great red spot? a long-lived, high-pressure storm on jupiter a place where reddish parti
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<span><u><em>A</em></u><em><u></u></em><em><u></u></em><u><em> long-lived, high-pressure storm on jupiter a place where reddish particles from io impact jupiter's surface</em></u></span><u><em> </em></u>
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The mass of an object on earth is 20 kg what will be the mass of that object on moon? And why? ​
Alla [95]

Answer:

20kg

Explanation:

Mass is a measure of the amount of matter in an object. The mass of an object, the amount of matter inside it does not change based on location. E.g. Objects do not lose matter when they travel to the moon.

Weight, on the other hand is the downward force you exert on the ground. Weight is calculated by multiplying the mass by the gravitational field strength and changes in different places with different gravitational strength. E.g. The moon's gravitational strength is 1/5 of Earth's so the mass of the object would stay the same but the weight would be only 20% of the weight is had on earth.

Hope this helped!

7 0
3 years ago
A ship maneuvers to within 2500 m of an island's 1800 m high mountain peak and fires a projectile at an enemy ship 610 m on the
Ne4ueva [31]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

t=(0-(250sin75)^2)/-9.8 
<span>the distance one is (2500+610)- (250m/s*cos75)*t=Dh Dh=horizontal distance </span>

<span>the max height one is d=0.5*9.8*t^2 </span>
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3 0
3 years ago
Read 2 more answers
A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.
erastova [34]

Complete Question

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).

J

(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)

MW

Answer:

The heat transferred is  Q = 5.866 * 10^9 J

The power is  P = 5866\  MW

Explanation:

From the question we are told that

      Mass of the water per second is m = 1917 \ kg

      The initial temperature of the water is T_i  = 35^oC

      The boiling point of water is  T_b = 100^oC

      The final temperature T_f = 450^oC

      The latent heat of vapourization of water is  c__{L}} = 2256*10^3 J/kg

      The specific heat of water c_w = 4184 J/kg^oC

      The specific heat of stem is C_s =1520 \ J/kg ^oC

Generally the heat needed each second is mathematically represented as

         Q = m[c_w (T_i - T_b) + m* c__{L}}  + m* c__{S}} (T_f - T_b)]

Then substituting the value

        Q = m[c_w [T_i - T_b] + c__{L}}  + C__{S}} [T_f - T_b]]

         Q = 1917 [(4184) [100 - 35] + [2256 * 10^3]  +[1520]  [450 - 100]]

         Q = 1917 * [3.05996 * 10^6]

         Q = 5.866 * 10^9 J

The power required is mathematically represented as

         P = \frac{Q}{t}

From the question t = 1\ s

So  

        P = \frac{5.866 *10^9}{1}

        P = 5866*10^6 \ W

        P = 5866\  MW

6 0
3 years ago
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