Is it true or false that cnidarians help build up coral reefs

Fuel cells have been developed that can generate a large amount of energy. For example, a hydrogen fuel cell works by combining

maksim [4K]

Electric power is given by:
Power = voltage x current
10000 = 15.8V
V = 633 Volts

7 0

4 years ago

Tides describe the regular rising and falling of ocean water. Tides are caused by A. earthquakes on the ocean floor. B. large st

Nadya [2.5K]

C. the gravitational pull of the Sun and Moon

5 0

3 years ago

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Please help. Questions are in the image.

yuradex [85]

Speed: measure of how fast an object travels
Velocity: measure of how fast, and in what direction, an object travels
Distance: how far an object has traveled on some path
Position: where an object is located in some reference system
Displacement: the difference between an objects starting position and it’s ending position

8 0

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:

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

5 0

3 years ago

A long solenoid has a radius of 4.0 cm and has 800 turns/m. If the current in the solenoid is increasing at the rate of 3.0 A/s,

kolbaska11 [484]

Answer:

Explanation:

Given that,

Radius of solenoid R = 4cm = 0.04m

Turn per length is N/l = 800 turns/m

The rate at which current is increasing di/dt = 3 A/s

Induced electric field?

At r = 2.2cm=0.022m

µo = 4π × 10^-7 Wb/A•m

The magnetic field inside a solenoid is give as

B = µo•N•I

The value of electric field (E) can

only be a function of the distance r from the solenoid’s axis and it give as,

From gauss law

∮E•dA =qenc/εo

We can find the tangential component of the electric field from Faraday’s law

∮E•dl = −dΦB/dt

We choose the path to be a circle of radius r centered on the cylinder axis. Because all the requested radii are inside the solenoid, the flux-area is the entire πr² area within the loop.

E∮dl = −d/dt •(πr²B)

2πrE = −πr²dB/dt

2πrE = −πr² d/dt(µo•N•I)

2πrE = −πr² × µo•N•dI/dt

Divide both sides by 2πr

E =- ½ r•µo•N•dI/dt

Now, substituting the given data

E = -½ × 0.022 × 4π ×10^-7 × 800 × 3

E = —3.32 × 10^-5 V/m

E = —33.2 µV/m

The magnitude of the electric field at a point 2.2 cm from the solenoid axis is 33.2 µV/m

where the negative sign denotes counter-clockwise electric field when looking along the direction of the solenoid’s magnetic field.

3 0

3 years ago

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