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3 years ago

SUUDVA

Physics

1 answer:

inn [45]3 years ago

5 0

3. The magnetic field is the area around a magnet in which there is magnetic force.

(1) Single touch method (2) Double touch method (3) Using electric current.

4. The difference between a permanent magnet and a temporary magnet is in their atomic structures. Permanent magnets have their atoms aligned all the time. Temporary magnets have their atoms aligned only while under the influence of a strong external magnetic field.

Electromagnets are temporary and artificial magnets. They are magnets that are only magnetic when there is a coil of wire with electricity running through it.

Motors and generators, Transformers, Relays, Electric bells and buzzers, Loudspeakers and headphones, Actuators such as valves, Magnetic recording and data storage equipment: tape recorders, VCRs, hard disks, MRI machines etc. (there is a lot)

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Pls Answer Quick Will Mark 5 Stars And brain lists

ra1l [238]

Answer:

Food mixer will convert electrical energy into kinetic energy

Explanation:

Energy conservation means that energy can not be created nor destroyed it can be converted from one form into another

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3 years ago

The Venn diagram compares protons with neutrons. Which shared property belongs in the region marked "B"?

telo118 [61]

is there any choices?

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3 years ago

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Un caballo parte del reposo y alcanza una velocidad de 15m/s en un tiempo de 8s. ¿ Cuál fue su aceleración y que distancia recor

Lemur [1.5K]

Answer: A 120 metros por segundo

Explanation: multiplicas la velocidad por el tiempo

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3 years ago

A car with 2 × 10^3 kg moving at a speed of 10 m/s collides and sticks with car B of mass of 3 × 10^3 kg initially at rest. How

stepan [7]

Answer:

$6 \times 10^4 \; \rm J$ .

Explanation:

KE lost = Total KE before Collision - Total KE after Collision.

For each car, the KE before collision can simply be found with the equation:

$\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2$ , where

$m$ is the mass of the car, and
$v$ is the speed of the car.

The $2 \times 10^3\; \rm kg$ car would have an initial KE of:

$\displaystyle \frac{1}{2} \times 2 \times 10^3 \times 10^2 = 10^5\; \rm J$ .

The $3 \times 10^3\; \rm kg$ car was initially not moving. Hence, its speed and kinetic energy would zero before the collision.

To find the velocity of the two cars after the collision, apply the conservation of momentum.

The momentum $p$ of an object is equal to its mass $m$ times its velocity $v$ . In other words, $p = m\cdot v$ .

Let the mass of the two cars be denoted as $m_1$ and $m_2$ , and their initial speeds $v_1$ and $v_2$ . Since the two cars are stuck to each other after the collision, their final speeds would be the same. Let that speed be denotes as $v_3$ .

Initial momentum of the two-car system:

$\begin{aligned}& m_1 \cdot v_1 + m_2 \cdot v_2 \\ &= 2 \times 10^3 \times 10 + 3 \times 10^3 \times 0 \\ &= 2 \times 10^4\; \rm kg \cdot m \cdot s^{-1}\end{aligned}$ .

After the collision, both car would have a velocity of $v_3$ (since they were stuck to each other.) As a result, the final momentum of the two-car system would be:

$m_1\cdot v_3 + m_2 \cdot v_3 = (m_1 + m_2)\, v_3$ .

Since momentum is conserved during the collision, the momentum of the system after the collision would also be $2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}$ . That is: $(m_1 + m_2) \, v_3 = 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}$ .

Solve for $v_3$ :

$\begin{aligned} v_3 &= \frac{(m_1 + m_2)\, v_3}{m_1 + m_2} \\ &= \frac{2 \times 10^4}{2 \times 10^3 + 3 \times 10^3} \\ &= \frac{2 \times 10^4}{5 \times 10^3} \\ &= 4 \; \rm m \cdot s^{-1}\end{aligned}$ .

Hence, the total kinetic energy after the collision would be:

$\begin{aligned} &\frac{1}{2}\, m_1 \, v^2 + \frac{1}{2}\, m_2\, v^2 \\ &= \frac{1}{2}\, (m_1 + m_2)\, v^2 \\ &= \frac{1}{2} \times \left(2 \times 10^3 + 3 \times 10^3\right) \times 4^2 \\ &= 4 \times 10^4\; \rm J\end{aligned}$ .

The amount of kinetic energy lost during the collision would be:

$\begin{aligned}&\text{KE After Collision} - \text{KE Before Collision} \\ &= 10^5 - 4 \times 10^4 \\&= 6\times 10^4\; \rm J \end{aligned}$ .

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3 years ago

A snowstorm was predicted in Chicago. Identify the possible upper air temperature, surface temperature, and air pressure of Chic

OverLord2011 [107]

The correct answer for this question is this one:

<span>A snowstorm was predicted in Chicago. The possible upper air temperature, surface temperature, and air pressure of Chicago on that day. Normal atmospheric pressure is 29.9 inches of mercury. </span><em>I'm pretty sure the answer is 40 for upper air, 29 for surface temp, and 30 for air pressure. </em>Hope this helps answer your question and have a nice day ahead.

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3 years ago

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