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solong [7]
3 years ago
7

According to the graph of displacement vs. time, what is the object's displacement at time = 60 s?

Physics
2 answers:
Paladinen [302]3 years ago
7 0

Answer:

The objects displacement is 0.0m.

Explanation:

krek1111 [17]3 years ago
4 0

Answer:

The straight line that is obtained, intercept it on the y-axis and the value of displacement will obtained.

Explanation:

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Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer dista
nadezda [96]

Distance = (speed) x (time)

Car A: Distance = (8 m/s) x (43 s)  =  344 meters

Car B: Distance = (7 m/s) x (50 s)  =  350 meters

350 meters is a longer distance than 344 meters.

<em>Car-B traveled a longer distance</em> than Car-A did.

5 0
2 years ago
Read 2 more answers
Suppose that two pith balls, each with a charge of 1 Coulomb, are hanging in the lab room, seperated by a distance of 1 meter, a
Soloha48 [4]
K (Q1) (Q2)/d^2 =

k : coulumb constant
Q1 : Charge on object 1
Q2 : Charge on object 2
d : distance between two charged object
4 0
3 years ago
The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separ
alexdok [17]

Answer:

U₁ = (ϵAV²)/6d

This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.

Explanation:

The energy stored in a capacitor is given by (1/2) (CV²)

Energy in the capacitor initially

U = CV²/2

V = voltage across the plates of the capacitor

C = capacitance of the capacitor

But the capacitance of a capacitor depends on the geometry of the capacitor is given by

C = ϵA/d

ϵ = Absolute permissivity of the dielectric material

A = Cross sectional Area of the capacitor

d = separation between the capacitor

So,

U = CV²/2

Substituting for C

U = ϵAV²/2d

Now, for U₁, the new distance between plates, d₁ = 3d

U₁ = ϵAV²/2d₁

U₁ = ϵAV²/(2(3d))

U₁ = (ϵAV²)/6d

This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.

3 0
3 years ago
A 14,700 N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force
Vlada [557]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

vo = 25 m/sec 
<span>vf = 0 m/sec </span>
<span>Fμ = 7100 N (Force due to friction) </span>
<span>Fg = 14700 N </span>
<span>With the force due to gravity, you can find the mass of the car: </span>
<span>F = ma </span>
<span>14700 N = m (9.8 m/sec²) </span>
<span>m = 1500 kg </span>
<span>Now, we can use the equation again to find the deacceleration due to friction: </span>
<span>F = ma </span>
<span>7100 N = (1500 kg) a </span>
<span>a = 4.73333333333 m/sec² </span>
<span>And now, we can use a velocity formula to find the distance traveled: </span>
<span>vf² = vo² + 2a∆d </span>
<span>0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d </span>
<span>0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d </span>
<span>-625 m²/sec² = (-9.466666666667 m/sec²) ∆d </span>
<span>∆d = 66.0211267605634 m </span>
<span>∆d = 66.02 m</span>
7 0
3 years ago
An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first
nordsb [41]

Answer:

a=0.555m/s^2

Explanation:

First we find the distance traveled from the moment the engineer reacts to the car, assuming uniform movement

X=VT

X=(18)(0.45)=8.1m

then we find the distance at which the deceleration begins, which is obtained by subtracting the total distance with the inner result

X=300-8.1=291.9

finally we use the equation for constant acceleration

Vf=0 final speed

Vo=18m/s= initial speed

X=291.9m

(Vf^2-Vo^2)/2X=a

(0-18^2)/(2*291.9)=a

a=0.555m/s^2

8 0
3 years ago
Read 2 more answers
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