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trapecia [35]
3 years ago
13

A skydiver falls 3km in 15s. How fast are they going. i need my answer in meters! (20 PTS)

Physics
1 answer:
Alchen [17]3 years ago
8 0
Assume the skydiver falls down straight.

Then, as you ask how fast he falls, the answer will be, the speed is not constant.

The speed is proportional with gravitational acceleration g and time passes by.

In other words, v = gt = 9.8t
You might be interested in
20. A photographer takes an average of 15 pictures per session. The total
fenix001 [56]

Answer:

A)

  • Independent variable -> number of sessions
  • Dependent variable -> number of pictures

B)

Values of the domain that make sense would be those from 0 on because there are no negative number of sessions; values of range that make sense would be also those from 0 on because there are no negative number of pictures.

C)

f(x)=15x, where "x" is the number of sessions.

If we have 22 session then:

f(22)=15*22; f(22)=f(22)=330

Explanation:

5 0
3 years ago
Humans impact the Earth in good AND bad ways. <br><br>A) True <br><br>B) False
const2013 [10]

Answer:

True

Explanation:

yes we can see that we are helping animals but we create pollution which is very bad

7 0
3 years ago
Why do you want the water to drip off the metal before it is placed in the calorimeter?
VMariaS [17]

Answer:

(A) The mass and the initial temperature of the calorimeter water will be incorrect and affect the calculation of the specific heat capacity of the metal.

8 0
3 years ago
A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is
lys-0071 [83]

Answer:

\omega_f=571.42\ rpm

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, a=1.5\ m/s^2

Displacement, d = 1.3 m

The angular acceleration is given by :

\alpha =\dfrac{a}{r}

\alpha =\dfrac{1.5}{0.033}

\alpha =45.46\ rad/s^2

The angular displacement is given by :

\theta=\dfrac{d}{r}

\theta=\dfrac{1.3}{0.033}

\theta=39.39\ rad

Using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

Here, \omega_i=0

\omega_f=\sqrt{2\alpha \theta}

\omega_f=\sqrt{2\times 45.46\times 39.39}

\omega_f=59.84\ rad/s

Since, 1 rad/s = 9.54 rpm

So,

\omega_f=571.42\ rpm

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0
3 years ago
A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is
malfutka [58]

Answer:  A) 6.38(10)^{6} m

Explanation:

The equation for the moment of inertia I of a sphere is:

I=\frac{2}{5}mr^{2} (1)

Where:

I=9.74(10)^{37}kg m^{2} is the moment of inertia of the planet (assumed with the shape of a sphere)

m=5.98(10)^{24}kg is the mass of the planet

r is the radius of the planet

Isolating r from (1):

r=\sqrt{\frac{5I}{2m}} (2)

Solving:

r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}} (3)

Finally:

r=6381149.077m \approx 6.38(10)^{6} m

Therefore, the correct option is A.

4 0
2 years ago
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