All categories

3 years ago

If 12V battery maintains a 4.5A current through a resistor what is the resistance of the resistor?

Physics

1 answer:

Dimas [21]3 years ago

5 0

Answer:

2.7ohms

Explanation:

Given parameters:

Voltage of the battery = 12V

Current = 4.5A

Unknown:

Resistance of the resistor = ?

Solution:

From Ohm's law, we know that;

V = IR

V is the voltage

I is the current

R is the resistance

So;

R = $\frac{V}{I}$ = $\frac{12}{4.5}$ = 2.7ohms

You might be interested in

Scientists can use spectral analysis of stars to determine which of the following?

RoseWind [281]

I do believe all of these but core elements can be determined by spectroscopy which includes the use of electromagnetic radiation. Both the surface and core temperature can be measured using light. Surface elements can be found because the absorption lines of different elements in the spectra of the star, but I haven't heard anything about using spectral analysis for core elements.

4 0

3 years ago

Read 2 more answers

A car battery seems to be malfunctioning (not providing the proper voltage and current to start your car). While the car is off

Stella [2.4K]

No. You cannot rule out the battery even after the open circuit voltage measurement. The open-circuit voltage may not have changed but the battery's internal resistance may have greatly increased.

4 0

3 years ago

In order to attain orbit around earth, the ATLAS-V rocket must accelerate up to a

IrinaK [193]

A) average acceleration = final velocity - initial velocity / time

= 7700 - 0 / 11

= 700ms^-2

B) force = mass x acceleration

= (3.05 x 105) x 700

= 320.25 x 700

= 224,175N

7 0

3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

4 years ago

A 60 kg man jumps down from a 0.8 m table. What is the speed when he

Rom4ik [11]

Answer: Speed = 4 m/s

Explanation:

The parameters given are

Mass M = 60 kg

Height h = 0.8 m

Acceleration due to gravity g= 10 m/s2

Before the man jumps, he will be experiencing potential energy at the top of the table.

P.E = mgh

Substitute all the parameters into the formula

P.E = 60 × 9.8 × 0.8

P.E = 470.4 J

As he jumped from the table and hit the ground, the whole P.E will be converted to kinetic energy according to conservative of energy.

When hitting the ground,

K.E = P.E

Where K.E = 1/2mv^2

Substitute m and 470.4 into the formula

470.4 = 1/2 × 60 × V^2

V^2 = 470.4/30

V^2 = 15.68

V = square root (15.68)

V = 3.959 m/s

Therefore, the speed of the man when hitting the ground is approximately 4 m/s

4 0

3 years ago