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lesya692 [45]
3 years ago
10

Give an example of a collision in real life. Use the law of conservation of energy to describe the transfer of momentum. Be sure

and discuss the momentum before and after the collision occurs. You will need at least 3 sentences to thoroughly answer this question.
Physics
1 answer:
Leno4ka [110]3 years ago
5 0

Momentum can be explained by a game of pool. A ball which is hit by the rod moves with certain mass and certain velocity. That ball hits another one and makes it move, also. So by giving it acceleration, a transfer of momentum is made.

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Please help with this and explain it,if you can.
vladimir2022 [97]

Answer:

displacement at 45 s  =  30

                           65 s  = 50

So the average speed over the interval from 45 s to 65 s is

(50 - 30) cm / 20 s = 1 cm / sec

As a check an average speed of 1 cm / sec for 20 sec will produce a

displacement of 1 cm / sec * 20 sec = 20 cm  or from 30 to 50 cm

4 0
3 years ago
An object has a mass of 5.20g. The mass of the object in kg is​
vampirchik [111]

Answer:

0.0520kg

Explanation:

divide by 1000

8 0
3 years ago
Weight of 9.81 N calculate its mass??
Fudgin [204]

Answer:

0.981kg

Explanation:

As we know,

1kg =10N

So,

mass = 9.81/10 = 0.981Kg

6 0
2 years ago
A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr
rjkz [21]

Answer:

<u>Part A</u>

I = 1.4 mW/m²  

<u>Part B</u>

β = 91.46 dB

Explanation:

<u>Part A</u>

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

                                            I = \frac{P}{A}

                                            I = \frac{P}{4\pi r^{2}}

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = 0.0013975 W/m²

                                 ≈  I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

                                     I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

<u>Part B</u>

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value.  It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;  

                                    β = 10log_{10}\frac{I}{I_{0}}  dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

           ∴        β = 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}

                      β = 10log_{10} (1.4 * 10^{9})

                      β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.                

4 0
3 years ago
A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and
Nastasia [14]

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

W=mg

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

\rho_a = 1.29 kg/m^3 is the air density

V=329 m^3 is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N

(c) 1524 N

The mass of the helium gas inside the balloon is

m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg

where \rho_h is the helium density; so we the total mass of the balloon+helium gas inside is

m=m_h+m_b=59 kg+210 kg=269 kg

So now we can find the weight of the balloon:

W=mg=(269 kg)(9.8 m/s^2)=2635 N

And so, the net force on the balloon is

F=B-W=4159 N-2635 N=1524 N

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

W'=(m'+m)g=B

where m' is the additional mass. Re-arranging the equation for m', we find

m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

B=\rho_a V g

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

4 0
3 years ago
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