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Marizza181 [45]

1 year ago

9

If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit willa) decrease

to one half of what it wasb) decrease to one fourth of what it wasc) quadruple in magnituded) double in magnitude

1 answer:

slega [8]1 year ago

5 0

Answer:

c) quadruple in magnitude

Explanation:

The power dissipated in the circuit is given by:

$P=I^2 R$

where

I is the current in the circuit

R is the total resistance of the circuit

In this problem:

- The current is doubled: I' = 2 I

- The resistance is kept constant: R' = R

So, the power dissipated is

$P' = (I')^2 R' = (2I)^2 R=4 I^2 R=4 P$

so, the power dissipated increase by a factor 4 (quadruples).

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antiseptic1488 [7]

The question with the complete options:

Look at these two sentences about Undeposited Funds. 1. By posting to Undeposited Funds, you can create a single bank deposit for multiple payments, making it easy ___________. 2. When receiving a payment, make sure _________________. Which of the options below correctly fills in the blanks? A.)1. To match your bank register with your bank statement; 2. the Deposit to account is Undeposited Funds

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4 0

1 year ago

A 4500-kg spaceship is in a circular orbit 190 km above the surface of Earth. It needs to be moved into a higher circular orbit

Maslowich

Answer:

Explanation:

Total energy of a satellite in an orbit , h height away

= - GMm /2 ( R + h )

When h = 380 km

Total energy of a satellite = $\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+380)\times10^3}$

= - 13.25 x 10¹⁰ J

When h = 190 km

Total energy of a satellite =

$\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+190)\times10^3}$

= - 13.63 x 10¹⁰ J

Diff

= 38 x 10⁸ J Energy will be required.

8 0

2 years ago

A blue puck has a velocity of 0i – 3j m/s and a mass of 4 kg. A gold puck has a velocity of 12i – 5j m/s and a mass of 6 kg. Wha

Mnenie [13.5K]

By definition, the kinetic energy is given by:
K = (1/2) * m * v ^ 2
where
m = mass
v = speed
We must then find the speed of both objects:
blue puck
v = root ((0) ^ 2 + (- 3) ^ 2) = 3
gold puck
v = root ((12) ^ 2 + (- 5) ^ 2) = 13
Then, the kinetic energy of the system will be:
K = (1/2) * m1 * v1 ^ 2 + (1/2) * m2 * v2 ^ 2
K = (1/2) * (4) * (3 ^ 2) + (1/2) * (6) * (13 ^ 2)
K = <span> 525</span> J
answer
The kinetic energy of the system is<span> <span>525 </span></span>J

6 0

1 year ago

If a single constant force acts on an object that moves on a straight line, the object's velocity is a linear function of time.

olya-2409 [2.1K]

Answer:

$F=mkv$

Explanation:

Given that

$v = v_i - kx$

We know that acceleration a given as

$a=\dfrac{dv}{dt}$

$v = v_i - kx$

$\dfrac{dv}{dt}=\dfrac{dv_i}{dt}-k\dfrac{dx}{dt}$

$\dfrac{dv}{dt}=0-k\dfrac{dx}{dt}$

We know that

$F=m\dfrac{dv}{dt}$

$F=-mk\dfrac{dx}{dt}$

$F=-mkv$

So the magnitude of force F

$F=mkv$

5 0

2 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0

1 year ago