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Ahat [919]
3 years ago
9

what type of motion occurs when an object spins around an axis without alternating its linear position

Physics
1 answer:
Zolol [24]3 years ago
4 0
Rotational motion occurs when an object spins around an axis without alternating its linear position.


hope this helped :)
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A quick USB charger claims its output current is 1.97Amp. We know that the standard USB output voltage is 5V. What is the output
Arte-miy333 [17]

Answer:

Output power of the charger is 9.85 watts.

Explanation:

It is given that,

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We need to find the output power of the charger. It can be determined using the following formula as :

P = V × I

P=5\ V\times 1.97\ A

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The output power of the charger is 9.85 watts. Hence, this is the required solution.

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3 years ago
What is the definition of chemical change
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3 0
3 years ago
A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget
mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, m_{1} = 4 kg

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v_{2} = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2

By plugging in the values we get,

KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)

KE = 40J + KE(lost)

Since,

Initial Kinetic energy = Final kinetic energy

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K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0
3 years ago
Recall that the differential equation for the instantaneous charge q(t) on the capacitor in an lrc-series circuit is l d 2q dt 2
Aloiza [94]
DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t)  by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150
5 0
3 years ago
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