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Anettt [7]
3 years ago
9

A 90-kg skydiver jumps from a height of 6000 m above the ground, falling head-first (pike position). The area of the diver is 0.

14 m2. The value for C is 1.0, and the density of air is 1.21 kg/m3. Determine the terminal velocity.
Physics
1 answer:
zzz [600]3 years ago
5 0

Answer:

102.097252061 m/s

Explanation:

F = Force

m = Mass = 90 kg

g = Acceleration due to gravity = 9.81 m/s² = a

C = Drag coefficient = 1

ρ = Density of air = 1.21 kg/m³

A = Surface area = 0.14 m²

v = Terminal velocity

Force is given by

F = ma

F=\dfrac{1}{2}\rho CAv^2\\\Rightarrow ma=\dfrac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\dfrac{ma}{\rho CA}}\\\Rightarrow v=\sqrt{2\dfrac{90\times 9.81}{1.21\times 1\times 0.14}}\\\Rightarrow v=102.097252061\ m/s

Terminal velocity of the skydiver is 102.097252061 m/s

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A block of mass m=1kg sliding along a rough horizontal surface is traveling at a speed v0=2m/s when it strikes a massless spring
topjm [15]

Here we can use the work energy theorem

W_f + W_s = K_f - K_i

here we know that

K_f = 0

as it come to rest finally

K_i = \frac{1}{2}mv_i^2

K_i = \frac{1}{2}\times 1\times 2^2

K_i = 2 J

now work done by friction force will be given as

W_f = - F_f \times d = -\mu mg d

W_f = - \mu(1)(9.8)(0.10) = - 0.98\mu

Work done by spring force is given as

W_s = \frac{1}{2}k(x_i^2 - x_f^2)

W_s = \frac{1}{2}(10)( 0 - 0.10^2)

W_s = -0.05 J

so now plug in all data above

- 0.05 - \mu(0.98) = 0 - 2

\mu = 1.99

so above is the friction coefficient


4 0
4 years ago
Nathan drops marbles down two ramps that have different lengths. It takes the marbles 10 seconds to reach the bottom of both ram
Sonbull [250]
<h2><u>Full Question:</u></h2>

Nathan drops marbles down two ramps that have different lengths. It takes the marbles 10 seconds to reach the bottom of both ramps.Which statement is TRUE?

answer choices

Marble 1 has a faster speed than Marble 2.

Marble 2 has a faster speed than Marble 1.

Both the marbles travel at the same speed.

There is not enough data to compare the speeds of marbles.

<h2><u>Answer:</u></h2>

Marble 2 has a faster speed than Marble 1.

Option B.

<h3><u>Explanation:</u></h3>

The speed is defined as the distance covered per unit time. Here in the question, 2 balls cover equal distances in same time.

Time taken by the ball = 10 seconds.

Distance covered by 1st ball = 20 cm.

Distance covered by 2nd ball = 3cm.

So speed of the 1st ball = 2cm/sec.

Speed of the 2nd ball = 3 cm /sec.

So,it's very much evident that speed of 2nd Marble is much higher than the speed of the 1st marble.

5 0
3 years ago
How can you use graphs to calculate the displacement of an object?
tekilochka [14]

Answer:

c. find the slope of the velocity time graph

8 0
3 years ago
Dwayne ‘The Rock’ Johnson needs to escape from the fourth floor of a burning building (in a movie). He ties a rope around his wa
ZanzabumX [31]

Answer:

Final Speed of Dwayne 'The Rock' Johnson = 15.812 m/s

Explanation:

Let's start out with finding the force acting downwards because of the mass of 'The Rock':

Dwayne 'The Rock' Johnson: 118kg x 9.81m/s = 1157.58 N

Now the problem also states that the kinetic friction of the desk in this problem is 370 N

Since the pulley is smooth, the weight of Dwayne Johnson being transferred fully, and pulls the desk with a force of 1157.58 N. The frictional force of the desk is resisting this motion by a force of 370 N. Subtracting both forces we get the resultant force on the desk to be: 1157.58 - 370 = 787.58 N

Now lets use F = ma to calculate for the acceleration of the desk:

787.58 = 63 x acceleration

acceleration = 12.501 m/s

Finally, we can use the motion equation:

v^2 - u^2 = 2*a*s

here u = 0 m/s (since initial speed of the desk is 0)

a = 12.501 m/s

and s = 10 m

Solving this we get:

v^2 - 0 = 2 * 12.501 * 10

v = 15.812 m/s

Since the desk and Mr. Dwayne Johnson are connected by a taught rope, they are travelling at the same speed. Thus, Dwayne also travels at            15.812 m/s when the desk reaches the window.

5 0
3 years ago
Objective lenses are contained in a ____________ that can be turned to put a particular objective lens in place to be used.
Masteriza [31]

Answer:

Revolving nosepiece

Explanation:

The revolving nosepiece is one of the parts of a microscope, used for holding the objective lenses. They can be turned to put a particular objective lens in place to be used in order to vary magnification.

6 0
3 years ago
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