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Softa [21]
3 years ago
5

Calculate the power if the current in a circuit is 3A and the resistance of the circuit is 50hms

Physics
1 answer:
Tresset [83]3 years ago
3 0

Answer:

450W

Explanation:

V= I*R

V=3A*50 Ohms

V=150V

P=I*V

P=3A*150V

P=450W

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A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2
Rudiy27

Answer:

Approximately 7.66\; \rm m.

Explanation:

<h3>Solve this question with a speed-time plot</h3>

The skateboarder started with an initial speed of u = 1.75\; \rm m \cdot s^{-1} and came to a stop when her speed became v = 0\; \rm m \cdot s^{-1}. How much time would that take if her acceleration is a = -0.20\; \rm m \cdot s^{-1}?

\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}.

Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:

\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m.

In other words, the skateboarder travelled 15.3\; \rm m up the slope until she came to a stop.

<h3>Solve this question with an SUVAT equation</h3>

A more general equation for this kind of motion is:

\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a},

where:

  • u and v are the initial and final velocity of the object,
  • a is the constant acceleration that changed the velocity of this object from u to v, and
  • x is the distance that this object travelled while its velocity changed from u to v.

For the skateboarder in this question:

\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}.

6 0
4 years ago
Identify the arrows that show input force
satela [25.4K]

The answer is that the reverse arrows pointing to the left needs to release energy. The molecules in the gas state are free such that they can travel from one point to another easily. They have the highest amount of energy. So, if you want the molecules to come closer together, you need to remove the energy to keep them in place. Therefore, the arrows pointing to the right require removal of energy.
8 0
3 years ago
An observer in frame S sees lightning simultaneously strike two points 100 m apart. The first strike occurs at xx1 = yy1 = zz1 =
Veseljchak [2.6K]

Answer:

a) 0, = -0.33 us

b) 140m

c) No, The event are not simultaneous i.e they did not occur at the same time, the second even (-0.33 usec) occurs 0.33 usec earlier than the first event.

Explanation:

a)

the lorentz factor expression is written as;

y = 1₀ / √(1 - (v²/c²))

where v  is the relative speed of an observer and c is the speed of light

so we were given that relative speed to be o.7c

therefore

y = 1 / √(1 - ((0.7c)² / c²))

y = 1 / √(1 - (0.49c² / c²))

y = 1 / √(1 - 0.49)

y = 1 / 0.7141

y = 1.4

1 - the coordinates  of the first event, the s' frame of reference is,

x1 ' = y(x1 - vt1) = 0

y1 ' = y1, z1' = z1 and

t1 ' = y [t1 - v/c²x1]

= 0

2 - the coordinates of the second event, the s ' frame of reference is'

x2 ' = y(x2-vt2)

= 1.4(100m - 0)

= 140m

y2 ' = y2, z2 ' = z2

t2 ' = y [ t2 - v/c²x2 ]

= 1.4 [ 0 - 0.7c/c²(100) ]

using speed of light c as 3*10^8

1.4 [ 0 - (0.7*3*10^8) / (3*10^8)²(100) ]

= -0.33 us

b)

distance between

delltaX' = X2' - X1'

= 140m - 0

= 140m

c)

No, The event are not simultaneous i.e they did not occur at the same time.

the second even (-0.33 us) occurs 0.33 us earlier than the first event.

4 0
3 years ago
What is the height of a building that an object is dropped from if it has a mass of 3 kg and hits the ground with a velocity of
Marizza181 [45]

Answer:

125 m

Explanation:

m = 3kg

v = 50m/s

u = 0m/s

a = +g = 10m/s²

s = H = ?

using the formula,

v² = u² + 2as

50² = 0² + 2(10)(H)

2500 = 20H

H = 2500/20

H = 125m

8 0
3 years ago
What is the correct answer?
bija089 [108]

Answer:

(B) 39

Explanation:

Use synthetic division:

-2 | 3  0  -1    3  1

____<u>-6 12 -22 38</u>

     3 -6 11  -19 | 39

This means that the remainder is 39

4 0
3 years ago
Read 2 more answers
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