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givi [52]
3 years ago
10

What is the power of a parallel circuit with a resistance of 1,000 and a current of 0.03 A?

Physics
1 answer:
Kobotan [32]3 years ago
7 0

i squared r = 0.03x0.03x1000=3x0.03x10=.9W

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7. A 50 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is over the bar is 1.0 m/s. Neglect air resist
luda_lava [24]

Answer:  5.05 m

Explanation:

4 0
3 years ago
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1. How many paths through which charge can flow would be shown in a diagram of a series
Elanso [62]
1. One
2. Oohm


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5 0
2 years ago
Over a short interval the coordinate of a car in the meters isgiven by x(t) = 27t - 4.0 t3 where time t is in seconds,at the end
wariber [46]

Answer:

5. -24 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s².

mathematically,

a = dv/dt ............................ Equation 1

Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.

But

v = dx(t)/dt

Where,

x(t) = 27t-4.0t³...................... Equation 2

Therefore, differentiating equation 2 with respect to time.

v = dx(t)/dt = 27-12t²............. Equation 3.

Also differentiating equation 3 with respect to time,

a = dv/dt = -24t

a = -24t .................... Equation 4

from the question,

At the end of 1.0 s,

a = -24(1)

a = -24 m/s².

Thus the acceleration = -24 m/s²

The right option is 5. -24 m/s²

4 0
2 years ago
Radio telescopes beam radio waves into space. To calculate the speed of the radio waves in space, which information, if any, is
Anastaziya [24]

Answer:

No more information is needed

Explanation:

Radio waves are electromagnetic energy, lower frequency forms of this type of energy that includes light and cosmic rays on the high frequency end that we are able to detect. So in free space (vacuum), radio waves travel at their fastest velocity, the “speed of light”. The reason for the quotation marks is because when light or radio waves are propagating through matter, we observe them traveling more slowly.

7 0
2 years ago
Desperate for help! please!!!!!!!!
lys-0071 [83]

0.4029 \mathrm{m} / \mathrm{s}^{2} is the acceleration of the box.

<u>Explanation:</u>

Given data:

Mass of the box = 3.74 kg

Flat friction-less ground is pulled forward by a 4.20 N force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle.

First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below.  Net horizontal force acting on the box (F) is given by

F=\left(4.20 \times \cos 50^{\circ}\right)+\left(2.25 \times \cos 122^{\circ}\right)

F=(4.20 \times 0.64278)+(2.25 \times-0.5299)

F = 2.699676 – 1.192275 = 1.507 N

Next, find acceleration of the box using Newton's second law of motion. This states that the link between mass (m) of an objects and the force (F) required to accelerate it. The equation can be given as

F=m \times acceleration\ (a)

\text {acceleration }(a)=\frac{F}{m}=\frac{1.507}{3.74}=0.4029 \mathrm{m} / \mathrm{s}^{2}

8 0
3 years ago
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