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german
3 years ago
12

What is meaning of convection​

Physics
1 answer:
crimeas [40]3 years ago
3 0

Answer:

Convection is heat transfer through the movement of liquids and gases.

You might be interested in
oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student
12345 [234]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0
3 years ago
At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0
3 years ago
In which region of the ear does resonance allow the brain to interpret sound answer
lakkis [162]
I'm not too sure what your asking but here are two answers that may help.
The ear drum amplifies the vibrations.
The cochlea changes vibrations into electric signals.
7 0
3 years ago
NEWTON LAWS
Vika [28.1K]

Answer:

it's Newton's first law of motion

As the law states that.

Everybody continues in its rest or of uniform motion unless an external force acts on it.

8 0
3 years ago
Read 2 more answers
A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how
Montano1993 [528]
<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity V_{o} has two components, because the brick was thrown at an angle \alpha=61\º:

V_{ox}=V_{o}cos\alpha   (1)

V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}  (2)

V_{oy}=V_{o}sin\alpha   (3)

V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}   (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

X=V_{ox}.t  (5)

Where:

X=10.1m is the distance where the brick landed

t is the time in seconds

If we already know X and V_{ox}, we have to find the time (we will need it for the following equation):

t= \frac{X}{ V_{ox}}  (6)

t=2.42s  (7)

<h2>In the y-axis: </h2>

-y=V_{oy}.t+\frac{1}{2}g.t^{2}   (8)

Where:

y is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

g=-9.8\frac{m}{s^{2}} is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}   (9)

-y=-10.52m   (10)

Multiplying by -1 each side of the equation:

y=10.52m >>>>This is the height of the building

3 0
3 years ago
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