Particles can enter the air by

A 5.4-kg ball is thrown into the air with an initial velocity of 35.2 m/s.a

kompoz [17]

The answer is

Ekin = 1/2 * m * v^2
Ekin = 1/2 * 5,4 * 35,2^2
Ekin = 3345,41 j (joule)

4 0

3 years ago

a coil has n turns enclosing an area of a. in a physics laboratory experiment, the coil is rotated during the time interval δt f

Mrac [35]

The correct answer for the question that is being presented above is this one:

Phi = BAsin(theta)
1. Phi(i) = BA 
2. Phi(f) = 0 
3. EMF = N(phi(i)-phi(f))/deltat

Here are the follow-up questions:

1. What is the total magnitude Phi_initial of the magnetic flux through the coil before it is rotated? 
2. What is the magnitude Phi_final of the total magnetic flux through the coil after it is rotated? 
3. What is the magnitude of the average emf induced in the coil?

8 0

3 years ago

Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some

erica [24]

If the density of water does not vary and the vents range in depth from about 1500 m to 3200 m below the surface, then the gauge pressure at a 2452-m deep vent is 224.268 atm.

Calculation:

Step-1:

It is given that the vents range in depth from about 1500 m to 3200 m below the surface. If we are assuming that the density of water does not vary. Then it is required to calculate the gauge pressure at a 2452-m deep vent.

The gauge pressure at a particular depth of ocean water is calculated as:

$P=\rho g h$

Here $\rho$ is the density of water, P is the required pressure, h is the depth of water, and g is the gravitational acceleration.

Step-2:

Now we are substituting the values to calculate the pressure at the depth of 2452-m.

$\\\begin{aligned}\\P&=\rho gh\\&=1030 (\text{ kg/m}^3)\times 9.8 (\text{ m/s}^2)\times 2452 \text{ m}\\&=24.75\times 10^6 \text{ Pa}\times\frac{1 \text{ atm}}{10.1325 \times10^4 \text{ Pa}}\\&=224.268 \text{ atm}\\\end{aligned}\\$

Learn more about gauge pressure here,

brainly.com/question/14012416

#SPJ4

5 0

2 years ago

When we see a meteor shower, it means that ________.

Irina18 [472]

Answer:

option A

Explanation:

The meteor shower is the celestial activity in which meteors are observed to radiate or originate from one point.

Meteors are nothing but dust or ice from the trails of comets. Most of the meteors are less than the size of the sand particle.

We will see comet shower when we earth will cross the orbit of the comet.

Hence, the correct answer is option A

6 0

4 years ago

At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m

eimsori [14]

Answer:

1) The time it takes the diver to move off the end of the diving board to the pool surface, $t_w$ , is approximately 1.392 seconds

2) The horizontal distance from the edge of the pool to where the diver enters the water, $d_w$ , is approximately 5.76 meters

Explanation:

1) The given parameters are;

The height of the diving board above the pool's surface, h = 9.5 m

The length by which the diving board over hangs the pool L = 2 m

The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s

Taking $t_w$ = The time it takes the diver to move off the end of the diving board to the pool surface

Therefore, we have from the equation of free fall;

h = 1/2 × g × $t_w$ ²

Where;

g = The acceleration due to gravity = 9.81 m/s²

Substituting the values, gives;

9.5 = 1/2 × 9.81 × $t_w$ ²

$t_w$ = √(9.5/(1/2 × 9.81)) ≈ 1.392 s

The time it takes the diver to move off the end of the diving board to the pool surface = $t_w$ ≈ 1.392 s

2) The horizontal distance, $d_w$ , in meters from the edge of the pool to where the diver enters the water is given as follows;

$d_w$ = L + v₀ × $t_w$ = 2 + 2.7× 1.392 ≈ 5.76 m

∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.

7 0

3 years ago