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Vlad [161]
4 years ago
6

Olivia is rolling down a hill on a skateboard and is accelerating at 2.4 m/s2. After 3 seconds, she runs into Reid at the bottom

of the hill. If her initial velocity was 1.5 m/s, what was her velocity when she hit Reid?
Physics
2 answers:
White raven [17]4 years ago
7 0

Answer:

8.7

Explanation:

A=2.4m/s2                        1.5m/s + (2.4m/s2)(3s)

Vi=1.5m/s                               1.5m/s + 7.2

Vf=?                                           Vf= 8.7m/s

T=3s

Sergeu [11.5K]4 years ago
5 0

did you ever get the answer?

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When an atom that has no charge looses two electrons it becomes a
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Answer: it becomes a positive ion

Explanation:

So, when an atom loses 2 electrons there will be no change in the number of neutrons. Therefore, an isotope will not form. Thus, it is concluded that when an atom with no charge loses two electrons, it becomes a positive ion.

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Europa, a satellite of Jupiter, is believed to have a liquid ocean of water (with a possibility of life) beneath its icy surface
Goryan [66]

Answer:

so maximum velocity for walk on the surface of europa is  0.950999 m/s

Explanation:

Given data

legs of length r =  0.68 m

diameter = 3100 km

mass = 4.8×10^22 kg

to find out

maximum velocity for walk on the surface of europa

solution

first we calculate radius that is

radius = d/2 = 3100 /2 = 1550 km

radius = 1550 × 10³ m

so we calculate no maximum velocity that is

max velocity = √(gr)    ...............1

here r is length of leg

we know g = GM/r²   from universal gravitational law

so G we know 6.67 × 10^{-11} N-m²/kg²

g = 6.67 × 10^{-11} ( 4.8×10^22 ) / ( 1550 × 10³ )

g = 1.33 m/s²

now

we put all value in equation 1

max velocity = √(1.33 × 0.68)

max velocity = 0.950999 m/s

so maximum velocity for walk on the surface of europa is  0.950999 m/s

3 0
3 years ago
A change in kinetic of an object is equal to the ___
podryga [215]

Answer:C..net work done on the object.

Explanation:

4 0
3 years ago
A V = 108-V source is connected in series with an R = 1.1-kΩ resistor and an L = 34-H inductor and the current is allowed to rea
soldi70 [24.7K]

Answer:

Explanation:

Given an RL circuit

A voltage source of.

V = 108V

A resistor of resistance

R = 1.1-kΩ = 1100 Ω

And inductor of inductance

L = 34 H

After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor

A. Time the inductor current will reduce to 12% of it's initial current

Let the initial charge current be Io

Then, final current is

I = 12% of Io

I = 0.12Io

I / Io = 0.12

The current in an inductor RL circuit is given as

I = Io ( 1—exp(-t/τ)

Where τ is time constant and it is given as

τ = L/R = 34/1100 = 0.03091A

So,

I = Io ( 1—exp(-t/τ))

I / Io = ( 1—exp(-t/τ))

Where I/Io = 0.12

0.12 = 1—exp(-t/τ)

0.12 — 1 = —exp(-t/τ)

-0.88 = -exp(-t/0.03091)

0.88 = exp(-t/0.03091)

Take In of both sides

In(0.88) = In(exp(-t/0.03091)

-0.12783 = -t/0.030901

t = -0.12783 × 0.030901

t = 3.95 × 10^-3 seconds

t = 3.95 ms

B. Energy stored in inductor is given as

U = ½Li²

So, the current at this time t = 3.95ms

I = Io ( 1—exp(-t/τ))

Where Io = V/R

Io = 108/1100 = 0.0982 A

Now,

I = Io ( 1—exp(-t/τ))

I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))

I = 0.0982(1—exp(-0.12783)

I = 0.0982 × 0.12

I = 0.01178

I = 11.78mA

Therefore,

U = ½Li²

U = ½ × 34 × 0.01178²

U = 2.36 × 10^-3 J

U = 2.36 mJ

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