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3 years ago

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Please can you help?? A skydiver is falling at terminal velocity when she opens her parachute. After a while she reaches a new t

erminal velocity. How does her new terminal velocity compare to her original terminal velocity?

1 answer:

Lubov Fominskaja [6]3 years ago

7 0

It will decrease because she is loosing momentum and she will begin to slow down.

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At a certain instant, the speedometer of the car indicates 80 km/h.

Scorpion4ik [409]

Answer:

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2 years ago

Which are functions of proteins?

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D. to be structural material

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3 years ago

Read 2 more answers

Analyze how some insects are able to move around on the surface of a lake or pond

koban [17]

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3 years ago

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

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3 years ago

In a charging process, 1 × 10^13 electrons are removed from one small metal sphere and placed on a second identical sphere. Init

liraira [26]

Answer:

r = 0.303m

= 30.3cm

Explanation:

Given that,

The number of electrons transferred from one sphere to the other,

n = 1 ×10 ¹³e le c t r o n s

The electrostatic potential energy between the spheres,

U = − 0.061 J

The charge on an electron,

q = − 1.6 × 10 ⁻¹⁹C

The coulomb constant,

K = 8.98755 × 10 ⁹ N ⋅ m ² / C 2²

Due to the transfer of electrons, both spheres become equally and oppositely.

The charge gained by the sphere due to the excess of the electron is:

q ₁ = n q

= 1 ×10 ¹³ * − 1.6 × 10 ⁻¹⁹

= -1.6 × 10⁻⁶C

The charge left on the first sphere is =

q ₂ = -q₁ = 1.6 × 10⁻⁶C

The electric potential energy between two point charges is given by the following equation:

U = K q ₁q ₂/r

q ₁ and q ₂ are the two charges.

r is the distance between the charge and the point.

K = 8.98755 × 10 ⁹ N ⋅ m ² / C ²

we have:

-0.061 = (8.98755 × 10 ⁹ * (-1.6 × 10⁻⁶)²) / r

r = (18.41 × 10 ⁻³) / 0.061

r = 0.303m

= 30.3cm

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3 years ago