Can I have some help

A 210 Ohm resistor uses 9.28 W of

IgorLugansk [536]

Resistance=R=210Ohm
Power=9.28W=P

Current=I

$\boxed{\sf P=I^2R}$

$\\ \sf\longmapsto I^2=\dfrac{P}{R}$

$\\ \sf\longmapsto I^2=\dfrac{9.28}{210}$

$\\ \sf\longmapsto I^2\approx0.04$

$\\ \sf\longmapsto I\approx\sqrt{0.04}$

$\\ \sf\longmapsto I\approx\sqrt{\dfrac{4}{100}}$

$\\ \sf\longmapsto I\approx\dfrac{\sqrt{4}}{\sqrt{100}}$

$\\ \sf\longmapsto I\approx\dfrac{2}{10}$

$\\ \sf\longmapsto I\approx0.2A$

4 0

2 years ago

A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.950 m. The sphere is swinging back and

Semenov [28]

Answer:

W = 0.842 J

Explanation:

To solve this exercise we can use the relationship between work and kinetic energy

W = ΔK

In this case the kinetic energy at point A is zero since the system is stopped

W = K_f (1)

now let's use conservation of energy

starting point. Highest point A

Em₀ = U = m g h

Final point. Lowest point B

Em_f = K = ½ m v²

energy is conserved

Em₀ = Em_f

mg h = K

to find the height let's use trigonometry

at point A

cos 35 = x / L

x = L cos 35

so at the height is

h = L - L cos 35

h = L (1-cos 35)

we substitute

K = m g L (1 -cos 35)

we substitute in equation 1

W = m g L (1 -cos 35)

let's calculate

W = 0.500 9.8 0.950 (1 - cos 35)

W = 0.842 J

7 0

3 years ago

What happens when two forces act in the same direction?

Wewaii [24]

Using the picture provided the forces are added together, because they are putting force on an object the same direction, thus the forces are added.

5 0

2 years ago

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A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

What is force?what is its si unit

uranmaximum [27]

Answer:

The pull or push which changes or tends to change the position of an object from rest to motion or motion to rest is known as force.

Its SI unit is newton(n).

8 0

2 years ago

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