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tiny-mole [99]
3 years ago
15

If a car travels 23.8 meters in 4.2 seconds, what is its speed?

Physics
1 answer:
Lena [83]3 years ago
5 0
Displacement = Initial Velocity * time + 0.5 * acceleration * time^2

Initial velocity * time = 0 since we assume that the car is launching from start without motion.

So, displacement = 0.5 * acceleration * time ^ 2

23.8 = 0.5 * acceleration * 4.2^2

23.8 = 0.5 * 17.64 * acceleration

23.8 = 8.82 * acceleration

23.8/8.82 = acceleration = 2.69 m/s^2

Final Velocity = initial velocity + acceleration * time

Final velocity = 0 + 2.69 * 4.2 = 11.3 m/s
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A dentist’s drill starts from rest. After 1.46 sof constant angular acceleration, it turns at arate of 27000 rev/min.Find the dr
Black_prince [1.1K]

Answer:

616.3 rad/s²

Explanation:

Given that

t= 1.46 s

Initial angular velocity ,ωi = 0 rad/s

Final angular velocity ωf= 27000 rev/min

Angular speed in the rad/s given as

\omega_f=\dfrac{2\pi N}{60}\ rad/s

Now by putting the values

\omega_f=\dfrac{2\times 27000}{60}\ rad/s

ωf=900 rad/s

We know that (if acceleration is constant)

ωf=ωi + α t

α=Angular acceleration

900 = 0 + α x 1.46

\alpha=\dfrac{900}{1.46}\ rad/s^2\\\alpha=616.43\ rad/s^2

Therefore the acceleration will be 616.3 rad/s²  

4 0
3 years ago
An overtone that's a whole number multiple of the fundamental frequency of a string is called a
alina1380 [7]
The correct answer in this case is B. Harmonic.

The remaining answers refer to other parts of a musical composition, and some such as the pitch can even be found in the study of human voice. 
6 0
3 years ago
If Siobhan hits a 0.25 kg volleyball with 0.5 N of force, what is the acceleration of the ball?
Alekssandra [29.7K]

Answer:

2 meters per second²

Explanation:

8 0
3 years ago
Read 2 more answers
The charges Q1=Q and Q2=4Q that are a distance d apart, repel each other with a force of 1.60 N. What would be the force between
gladu [14]

Answer:

50.4 N

Explanation:

Q1 = Q

Q2 = 4 Q

Distance = d

The force is given by

F = \frac{KQ_{1}Q_{2}}{d^{2}}

1.60 = \frac{4KQ^{2}}{d^{2}}    .... (1)

Now,

Q3 = 2 Q

Q4 = 7 Q

distance = d/3

F' = \frac{9KQ_{3}Q_{4}}{d^{2}}

F' = \frac{126KQ^{2}}{d^{2}}   .... (2)

Divide equation (2) by equation (1), we get

F' / 1.60 = 126 / 4

F' = 50.4 N

Thus, the force is 50.4 N.

7 0
3 years ago
A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge
Digiron [165]

Answer:

F = 2.49 x 10⁻⁹ N

Explanation:

The electrostatic force between two charged bodies is given by Colomb's Law:

F = \frac{kq_1q_2}{r^2}\\

where,

F = Electrostatic Force = ?

k = colomb's constant = 9 x 10⁹ N.m²/C²

q₁ = charge on proton = 1.6 x 10⁻¹⁹ C

q₂ = second charge = 1.4 C

r = distace between charges = 0.9 m

Therefore,

F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^{-19}\ C)(1.4\ C)}{(0.9\ m)^2}

<u>F = 2.49 x 10⁻⁹ N</u>

8 0
3 years ago
Read 2 more answers
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