All categories

3 years ago

Which of the following changes will result in a stronger electromagnet?

2 answers:

8 0

The correct option is C. i.e. Using more coils of wire

Explanation:

The strength of an electromagnet can be increased in the following ways:

1. By increasing the number of coils.
2. By increasing the current/voltage
3. Using an iron core, increases the strength of the electromagnet.

nirvana33 [79]3 years ago

7 0

The answer is going to be C

You might be interested in

How long will your trip take if you travel 4000 m at an average speed of 8m/s

SpyIntel [72]

Answer:

500 sec

8 min 20 sec

Explanation:

Hi there !

8 m ................ 1 s

4000 m ........ x s

x = 4000m×1s/8m = 500 sec = 8 min 20 sec

Good luck !

8 0

2 years ago

Read 2 more answers

An object of mass 2 kg starts from rest and is allowed to slide down a

WARRIOR [948]

Answer: B

Explanation:

Given that an object of mass 2 kg starts from rest and is allowed to slide down a frictionless incline so that its height changes by 20 m.

The parameters given from the question are:

Mass M = 2kg

Height h = 20m

Let g = 9.8m/s^2

At the bottom of the incline plane, the object will experience maximum kinetic energy.

From conservative of energy, maximum K.K.E = maximum P.E

Maximum P.E = mgh

Maximum P.E = 2 × 9.8 × 20 = 392 J

But

K.E = 1/2mv^2

Substitute the values of energy and mass into the formula

392 = 1/2 × 2 × V^2

V^2 = 392

V = sqrt( 392 )

V = 19.8 m/s

V = 20 m/s approximately

7 0

3 years ago

It’s upside down, I know.

Paladinen [302]

Where's the question?

3 0

3 years ago

What rapid changes to earths surface are caused by shifting plates

PtichkaEL [24]

I am not sure if this is the answer you are looking for but a earthquake occurs when the plates shift.
Hope this helped.

4 0

2 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

2 years ago