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2 years ago

How do the forces on a skydiver affect the velocity of the skydiver as they fall? Hellp

2 answers:

8 0

When a skydiver falls, he constantly accelerates. Therby, falling faster and faster. But also he is slowed down a bit due to the force of air resistance.

solmaris [256]2 years ago

3 0

Answer:

Explanation:

As a skydiver falls, he accelerates downwards, gaining speed with each second. The increase in speed is accompanied by an increase in air resistance. This force of air resistance counters the force of gravity.

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A child is swinging back and forth with a constant period and amplitude. Somewhere in front of the child, a stationary horn is e

Amanda [17]

Answer:

Explanation:

We shall apply concept of Doppler's effect of apparent frequency to this problem . Here observer is moving sometimes towards and sometimes away from the source . When observer moves towards the source , apparent frequency is more than real frequency and when the observer moves away from the source , apparent frequency is less than real frequency . The apparent frequency depends upon velocity of observer . The formula for apparent frequency when observer is going away is as follows .

f = f₀ ( V - v₀ ) / V , f is apparent , f₀ is real frequency , V is velocity of sound and v is velocity of observer .

f will be lowest when v₀ is highest .

velocity of observer is highest when he is at the equilibrium position or at middle point .

So apparent frequency is lowest when observer is at the middle point and going away from the source while swinging to and from before the source of sound .

3 0

2 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is $14,000 + 10,000x − 26,000x^{2}$ , where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

W = $\int_{0}^{0.54} F dx$

= $\int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx$

= $[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}$

= 7560 + 1458 - 1364.69

= 7653.31 J

or, = 7.65 kJ (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0

2 years ago

Plz help >:

svlad2 [7]

Answer:

10m

Explanation:

The object distance and image distance is the same from the mirror. so the image is 5m behind the mirror.

5+5=10

5 0

2 years ago

Calculating ph how is ph related to the concentration of hydronium ions. True or False

Dmitry_Shevchenko [17]

I think this is TRUE. Ph is calculating the acidic acids in water. And you need to know the concentration of hydronium ions. Hope this helped !

7 0

2 years ago

In the diagram, q1= +8.0 C, q2= +3.5 C, and q3 = -2.5 C. q1 to q2 is 0.10 m, q2 to q3 is 0.15 m. What is the net force on q2? La

yulyashka [42]

Answer:

f(t) = 28,7 [N]

Explanation: IMPORTANT NOTE: IN PROBLEM STATEMENT CHARGES ARE IN C (COULOMBS) AND IN THE DIAGRAM IN μC. WE ASSUME CHARGES ARE IN μC.

The net force on +q₂ is the sum of the force of +q₁ on +q₂ ( is a repulsion force since charges of equal sign repel each other ) and the force of -q₃ on +q₂ ( is an attraction force, opposite sign charges attract each other)

The two forces have the same direction to the right of charge q₂, we have to add them

Then

f(t) = f₁₂ + f₃₂

f₁₂ = K * ( q₁*q₂ ) / (0,1)²

q₁ = + 8 μC then q₁ = 8*10⁻⁶ C

q₂ = + 3,5 μC then q₂ = 3,5 *10⁻⁶ C

K = 9*10⁹ [ N*m² /C²]

f₁₂ = 9*10⁹ * 8*3,5*10⁻¹²/ 1*10⁻² [ N*m² /C²]* C*C/m²

f₁₂ = 252*10⁻¹ [N]

f₁₂ = 25,2 [N]

f₃₂ = 9*10⁹*3,5*10⁻⁶*2,5*10⁻⁶ /(0,15)²

f₃₂ = 78,75*10⁻³/ 2,25*10⁻²

f₃₂ = 35 *10⁻¹

f₃₂ = 3,5 [N]

f(t) = 28,7 [N]

5 0

2 years ago

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