Ask question

Ask question

All categories

3 years ago

9

A battery supplies E volts to a circuit containing a resistor R. How could the current in the resistor be increased? Select all

that apply.
Replace the resistor with a smaller resistor.
Replace the resistor with a larger resistor.
Connect a smaller battery.
Connect a larger battery.

1 answer:

anastassius [24]3 years ago

5 0

Answer:

I=V/R

so larger bsttery or small resistor can increase the current value.

You might be interested in

mechanical equivalent of heat Do the data for the second part of the experiment support or refute the second hypothesis

Ksivusya [100]

The data for the first part of the experiment support the first hypothesis.
As the force applied to the cart increased, the acceleration of the cart increased.
Since the increase in the applied force caused the increase in the cart's acceleration, force and acceleration are directly proportional to each other, which is in accordance with Newton's second law.

When we state something about the results on the basis whether the observed data supports the original hypothesis, we say that we are concluding the results.

What is the relationship between force and acceleration based on Newton's 2nd law?

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

Learn more about Newton's second law of motion brainly.com/question/13447525

#SPJ4

6 0

2 years ago

A non-reflective coating that has a thickness of 198 nm (n = 1.45) is deposited on top of a substrate of glass (n = 1.50). What

spin [16.1K]

Answer:

The wavelength is $\lambda_ 1 = 574.2 nm$

Explanation:

From the question we are told that

The thickness is $t = 198 nm = 198 *10^{-9 }\ m$

The refractive index of the non-reflective coating is $n_m = 1.45$

The refractive index of glass is $n_g = 1.50$

Generally the condition for destructive interference is mathematically represented as

$2 * n_m * t * cos (\theta) = n * \lambda$

Where $\thata$ $\theta$ is the angle of refraction which is 0° when the light is strongly transmitted

and n is the order maximum interference

so

$\lambda = \frac{2 * n * t * cos (\theta )}{n}$

at the point n = 1

$\lambda _1 = \frac{2 * 1.45 * 198*10^{-9} * cos (0 )}{1}$

$\lambda_1 = 574.2 *10^{-9}$

$\lambda_1 = 574.2 nm$

at n =2

$\lambda _2 = \frac{\lambda _1 }{2}$

$\lambda _2 = \frac{574.2*10^{-9} }{2}$

$\lambda _2 = 2.87 1 *10^{-9} \ m$

$\lambda _2 = 287. 1 nm$

Now we know that the wavelength range of visible light is between

$390 \ nm \to 700 \ nm$

So the wavelength of visible light that is been transmitted is

$\lambda_ 1 = 574.2 nm$

6 0

4 years ago

A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2

Leno4ka [110]

Answer:

A. T=15.54 °C

B. Q/A= 0.119 W/m2

Explanation:

To solve this problem we need to use the Fourier's law for thermal conduction:

$Q= kA\frac{dT}{dx}$

Here, the rate of flow per square meter must be the same through the complete wall. Therefore, we can use it to find the temperature at the plane where the wood meets the Styrofoam as follows:

$\frac{Q}{A} =\frac{T_1-T_0}{d_w}k_w=\frac{T_2-T_1}{d_s}k_s\\T_1(\frac{k_w}{d_w}+\frac{k_s}{d_s})=T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}\\T_1=\frac{T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}}{\frac{k_w}{d_w}+\frac{k_s}{d_s}}\\T_1= 15.54 \°C$

Then, to find the rate of heat flow per square meter, we have:

$\frac{Q}{A}=\frac{T_1-T_0}{d_w}k_w=0.119 \frac{W}{m^2}\\\frac{Q}{A}=\frac{T_2-T_1}{d_s}k_s= 0.119 \frac{W}{m^2}$

$T_0: Temperature \ in \ the \ house\\T_1: Temperature \ at \ the \ plane \ between \ wood \ and \ styrofoam\\T_2: Temperature \ outside\\k_w: k \ for \ wood\\d_w: wood \ thickness\\k_s: k \ for \ styrofoam\\d_s: styrofoam \ thickness$

7 0

3 years ago

A light ray in water is incident on a boundary with air at an angle of 22°. The refractive index of water is 1.33 and that of ai

Komok [63]

The answer to this question is C

6 0

4 years ago

Read 2 more answers

At a certain distance from the center of the Earth, a 0.4-kg object has a weight of 2.0 N. (a) Find this distance. (b) If the ob

Alika [10]

Answer:

a) The distance of the object from the center of the Earth is 8.92x10⁶ m.

b) The initial acceleration of the object is 5 m/s².

Explanation:

a) The distance can be found using the equation of gravitational force:

$F = \frac{GMm}{r^{2}}$

Where:

G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²

M: is the Earth's mass = 5.97x10²⁴ kg

m: is the object's mass = 0.4 kg

F: is the force or the weight = 2.0 N

r: is the distance =?

The distance is:

$r = \sqrt{\frac{GMm}{F}} = \sqrt{\frac{6.67 \cdot 10^{-11} Nm^{2}/kg^{2}*5.97 \cdot 10^{24} kg*0.4 kg}{2.0 N}} = 8.92 \cdot 10^{6} m$

Hence, the distance of the object from the center of the Earth is 8.92x10⁶ m.

b) The initial acceleration of the object can be calculated knowing the weight:

$W = ma$

Where:

W: is the weight = 2 N

a: is the initial acceleration =?

$a = \frac{W}{m} = \frac{2 N}{0.4 kg} = 5 m/s^{2}$

Therefore, the initial acceleration of the object is 5 m/s².

I hope it helps you!

4 0

3 years ago