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2 years ago

Is an ocean wave electromagnetic or mechanical

Physics

1 answer:

swat322 years ago

3 0

Answer:

mechanical

Explanation:

the energy is carried by water <3 hope this helped

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What is the current in a circuit if 3.50 C of charge passes any given point in a circuit in 1.80 s?

stepan [7]

Answer:

I = 1.944 A

Explanation:

<u>Given the following data;</u>

Charge, Q = 3.5C

Time, t = 1.8 secs

Charge of a substance is given by the formula;

$Q = It$

Where;

Q is the amount of charge measured in Colombs.

I is the current measured in Amperes.

t is the time measured in seconds.

Making I the subject of formula, we have

$I = \frac {Q}{t}$

$I = \frac {3.5}{1.8}$

I = 1.944 A

Therefore, the current in the circuit is 1.944 Amperes.

7 0

3 years ago

Match the synthetic materials with the processes used to make them.

BlackZzzverrR [31]

We can’t see anything

3 0

3 years ago

A weightlifter does 450 J of work on a barbell in 3 s. How much power is the weightlifter generating?

astra-53 [7]

450 J / 3 s = 150 J/s = 150 watts.

8 0

4 years ago

An LP turntable must spin at 3.500 rad/s to play a record. How much torque must the motor deliver if the turntable is to reach i

zmey [24]

Answer:

0.00321 Nm

Explanation:

$\omega_f$ = Final angular velocity = 3.5 rad/s

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 2 rev

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{3.5^2-0^2}{2\times 2\pi \times 1.8}\\\Rightarrow \alpha=0.54156\ rad/s^2$

Moment of inertia is given by

$I=mr^2\\\Rightarrow I=0.25\times 0.154^2\\\Rightarrow I=0.005929\ kgm^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=0.005929\times 0.54156\\\Rightarrow \tau=0.00321\ Nm$

The torque required is 0.00321 Nm

8 0

4 years ago

I need help with this physics question.

Bumek [7]

The acceleration will increase by 61.3%.

Explanation:

The centripetal acceleration $a_c$ is given by

$a_c = \dfrac{v^2}{r}$

If the velocity of the object increases by 27.0%, then its new velocity v' becomes

$v' = 1.270v$

The new centripetal acceleration $a'_c$ becomes

$a'_c = \dfrac{(1.270v)^2}{r} = 1.613 \left(\dfrac{v^2}{r} \right)$

$\:\:\:\:\:\:\:\:\:= 1.613a_c$

8 0

3 years ago