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Alisiya [41]
3 years ago
15

Wel.

Physics
1 answer:
Law Incorporation [45]3 years ago
4 0

Answer:

1 is c 2 is a and 3 is b hope that helped!

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A car traveling at 60km/h undergoes uniform acceleration at a rate of 2/ms^2 until is velocity reached 120km/h determine the dis
jenyasd209 [6]

Explanation:

Given that,

Initial speed of a car, u = 60 km/h = 16.67 m/s

Acceleration, a = 2m/s²

Final speed, v = 120 km/h = 33.33 m/s

We need to find the distance traveled and the time taken to make the distance.

acceleration = rate of change of velocity

a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{33.33 -16.67 }{2}\\\\t=8.33\ s

let the distance be d.

d=\dfrac{v^2-u^2}{2a}\\\\d=\frac{33.33^{2}-16.67^{2}}{2(2)}\\\\d=208.25\ m

Hence, the distance traveled and the time taken to make the distance is 208.25 m and 8.33 seconds respectively.

3 0
2 years ago
You are on a ParKour course. First you climb a angled wall up 9.5 meters. They you shimmy along the edge of a 3.5 meter long wal
Svet_ta [14]

The average speed would be  0.65 m/s, therefore the correct option is (A).

The average speed is calculated by the formula

Average speed= (total distance/ total time)

The total distance of the trip=9.5+3.5+15=28 m

The total time of trip=43 sec

Therefore the average speed=28/43=0.65 m/s.


8 0
3 years ago
Light from the sun reaches Earth in 8.3min. The speed of light is 3.00 x 10^8 m/s. How far is the Earth from the sun?
strojnjashka [21]
The answer is this, but i don't know how to simplify it. 3x^100000000<span />
5 0
3 years ago
Read 2 more answers
In which medium would sound travel the fastest? a)across a room b)in a swimming pool c)through outer space d)through a railroad
lora16 [44]

d)through a railroad track

7 0
3 years ago
A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the
const2013 [10]

Answer:

The equation for the object's displacement is u(t)=0.583cos11.35t

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in

The angular speed is:

w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s

The damping force is:

F_{D} =cu

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in

The critical damping is equal:

c_{c} =\frac{2k}{w} =\frac{2*5.33}{11.35} =0.94lbs/in

Like cc>c the system is undamped

The equilibrium expression is:

u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t

3 0
3 years ago
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