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kiruha [24]
3 years ago
7

a student pushes glider on an air track with a force of 2.0 N for 4.0 seconds over a distance of 1.0 M. How much work did the st

udent do on the glider?
Physics
1 answer:
Agata [3.3K]3 years ago
5 0

Answer:

2.6

Explanation:

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Suppose the coefficient of kinetic friction between mA and the plane in the figure(Figure 1) is μk = 0.15, and that mA=mB=2.7kg.
luda_lava [24]
A ) 
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81  + 2.7 a
T = 26.487 + 2.7 a
26.487 + 2.7 a + 2.7 a - 2.7 · 9.81 · 0.574 = 0.15 · 2.7 · 9.81 · 0.819
5.4 a + 26.487 - 15.2023 = 3.2539
5.4 a = 8.0296
a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487 
26.487 - 15.2035 = (Mi) · 2.7 · 9.81 · 0.819
11.2835 = (Mi) · 21.69
(Mi) = 11.2835 : 21.69 = 0.52
4 0
3 years ago
If two asteroids moved closer together, what would be the result on the gravitational force each asteroid exerts on the other?
spayn [35]
Gravitational force depends on inverse square law. That is, gravitational force is inversely proportional to square of distance between asteroids.
As distance between them decreases, gravitational force increases. Hence A is correct.
7 0
3 years ago
Read 2 more answers
How long would it take to travel 10 light years at the speed of light?
Vladimir79 [104]

A light year is the DISTANCE light travels through vacuum in 1 year.

If light is traveling through vacuum, then it's traveling at the speed of light in vacuum. If a student at home at the beginning of the trip is holding the clock, then ...

Traveling 1 light year takes 1 year.

Traveling 2 light years takes 2 years.

Traveling 3 light years takes 3 years.

Traveling 10 light years takes 10 years.

If the light is traveling through some other substance, or if the clock is traveling along with the light, then these numbers all change.

YOU cannot travel at the speed of light. We have to just leave it at that

6 0
3 years ago
What is the current in a 160V circuit if the resistance is 2Ω?<br> V=<br> I=<br> R=
Alex787 [66]

Explanation:

v = IR

v= 160 R = 2

<u>160</u> = <u>2I</u>

2 2

I = 80A

4 0
2 years ago
Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
3 years ago
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