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Ilia_Sergeevich [38]
3 years ago
11

A car is traveling at some speed when it accelerates at 6 m/s2 for 3 seconds. If the car travels 39 meters in this time, how fas

t was it going before it sped up?
Physics
1 answer:
Alchen [17]3 years ago
6 0

Answer:

4 m/s

Explanation:

Given:

Δx = 39 m

a = 6 m/s²

t = 3 s

Find: v₀

Δx = v₀ t + ½ at²

39 m = v₀ (3 s) + ½ (6 m/s²) (3 s)²

v₀ = 4 m/s

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A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele
Fantom [35]
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, \Delta t:
I= \frac{Q}{\Delta t}
First we need to find the net charge flowing at a certain point of the wire in one second, \Delta t=1.0 s. Using I=0.92 A and re-arranging the previous equation, we find
Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C

Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}
3 0
3 years ago
An ice skater is spinning at 6.00 rev/s with his moment of inertia being 0.400 kg/m2. Calculate his new moment of inertia if he
labwork [276]

Answer:

New moment of inertia will be I=1.92kgm^2

Explanation:

It is given initially angular velocity \omega =6rev/sec=6\times 2\pi =37.68rad/sec

Moment of inertia I=0.4kgm^2

Angular momentum is equal to L=I\omega =37.68\times 0.4=15.072kgm^2/sec

Now angular velocity is decreases to \omega =1.25rev/sec=1.25\times 2\times 3.14=7.85rad/sec

As we know that angular momentum is conserved

So 15.072=I\times 7.85

I=1.92kgm^2

So new moment of inertia will be I=1.92kgm^2

4 0
4 years ago
If a car moves with a uniform speed of 2m/s in a circle of radius 0.4m, what is its angular speed?
REY [17]

Answer:

The car's angular speed is \frac{rad}{s}.

Explanation:

Angular velocity is usually measured with \frac{radians}{sec}, so I'm going to use that to answer your question.

The relationship between tangential velocity and angular velocity (ω)  is given by:

V = \omega R

Using the values from the question, we get:

2 \frac{m}{s} = \omega (0.4m)

\omega = 5 \frac{rad}{s}

Therefore, the car's angular speed is 5 \frac{rad}{s}.

Hope this helped!

3 0
3 years ago
A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is
Umnica [9.8K]

Answer:

<h2>Magnitude of the second charge is -4.17*10^{-7}C</h2>

Explanation:

According to columbs law;

F = kq1q2/r^{2}

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

0.03=9*10^{9}*  -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C

6 0
4 years ago
Read 2 more answers
By the time she managed/had managed to open the door, the postman already went/had already gone.
leva [86]

Answer:

<h3>Answer Sentence: </h3><h2><em><u>By the time she had managed to open the door, the postman had already gone</u></em>.</h2>
8 0
3 years ago
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