PLEASE HELP EASY MULTIPLE CHOICE!!!!!!!!!!!

Answer:

1.3 m/s

Explanation:

It is given that,

Mass of bird A, [tex]m_A=2.2\ kg[/tex]

Mass of bird B, [tex]m_B=1.7\ kg[/tex]

Initial speed of bird A is 0 as it was at rest

Initial speed of bird B is 3 m/s

We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,

[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]

So, the center of mass for this system is 1.3 m/s.

Answer:

It would change the amount of heat produced in the transmission line to four times the previous value.

Explanation:

Given;

initial voltage in the transmission line, V₁ = 500 kV = 500,000 V

Final voltage in the transmission line, V₂ = 1 MV = 1,000,000

The power lost in the transmission line due to heat is given by;

[tex]P = \frac{V^2}{R}[/tex]

Power lost in the first wire;

[tex]P_1 = \frac{V_1^2}{R}[/tex]

[tex]R = \frac{V_1^2}{P_1}[/tex]

Power lost in the second wire

[tex]P_2 = \frac{V_2^2}{R}\\\\ R = \frac{V_2^2}{P_2}[/tex]

Keeping the resistance constant, we will have the following equation;

[tex]\frac{V_2^2}{P_2} = \frac{V_1^2}{P_1} \\\\P_2 = \frac{V_2^2P_1}{V_1^2}\\\\[/tex]

[tex]P_2 = \frac{(1,000,000)^2P_1}{(500,000)^2}\\\\P_2 =4P_1[/tex]

Therefore, it would change the amount of heat produced in the transmission line to four times the previous value.

Answer:

50 m

Explanation:

The relationship between the intensity of sound in dB and distance is given by the formula:

[tex]B_2=B_1+20log(\frac{R_1}{R_2} )\\\\Where \ B_2\ is \ the\ sound\ intensity\ at\ distance\ R_2\ and\\B_1\ is \ the\ sound\ intensity\ at\ distance\ R_1\ \\\\Given\ that: B_1=100\ dB, R_1=5\ m, B_2=80\ dB\\\\B_2=B_1+20log(\frac{R_1}{R_2} )\\\\80=100+20log(\frac{5}{R_2} )\\\\-20=20log(\frac{5}{R_2} )\\\\log(\frac{5}{R_2} )=-1\\\\\frac{5}{R_2}=10^{-1}\\\\\frac{5}{R_2}=0.1\\\\R_2=5/0.1=50\ m[/tex]

Answer: It will be about 44.1m

Explanation:

Answer:

electrical energy transforming into sound energy in speaker

Answer:

the first one. Electrical energy transforming into sound energy in a speaker

Answer:

Explanation:

Find the complete question attached

Using the principle of moment

Clockwise moment = Anticlockwise moment

AntiClockwise moment = M × 2.0

ACW moment = 2M

Clockwise moment = 40×4

Clockwise moment = 160kgcm

Equate both expression and calculate M

2M = 160

M = 160/2

M = 80kg

Hence the mass of his friend is 80kg

Answer:

Covalent Bond

Explanation:

i took the test , mark me brainliest.

Answer: Electrical

Explanation: Atoms are tied together by electrical bonding forces.

Answer:

South pole

Explanation:

In a bar magnet, field lines go from the North Pole to the South Pole (outside the magnet).

As the earth magnetic field lines go from South Pole (geographic) to the North one, this means that the North pole (geographic) really behaves as a South Pole (magnetic).

Answer:

Explanation:

If the ray were not deviated, it would travel straight through the mirror. Due to the mirror, the incident ray is reflected at 30°. The ray travels 30° + 30° = 60°. The angle of deviation is 180° - 60° = 120°.

This question is incomplete, the complete question is;

The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"

Answer:

the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

Explanation:

Given that;

P₁ = 1.00 atm

P₂ = ?

V₁ = 1 L

V₂ = 1.60 L

the temperature of the gas is kept constant

we know that;

P₁V₁ = P₂V₂

so we substitute

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

Answer:

 Em₀ = U = m g h ,  Em_{f} = K = ½ m v²

Explanation:

When a car is on a ramp it has a certain amount of mechanical energy. At the highest point of the ramp the mechanical energy is fully potential given by

          Em₀ = U = m g h

As part of this energy descends down the ramp, part of this energy is transformed into kinetic energy and has one part of each, even though the sum remains the initial energy

            Em = K + U = ½ m v² + mg y

        y <h

when it reaches the bottom of the ramp it has no height therefore there is no potential energy, all of it has been transformed into kinetic energy

          Em_{f} = K = ½ m v²

This energy transformation is in the case that the friction force is zero.

If there is a friction force, it performs work against the low car, it is reflected in an increase in the internal energy (temperature) of the car. In this case the energy in the lower part is less than the initial one by a factor

         [tex]W_{nc}[/tex] = - fr L

therefore the numeraire values ​​of the velocity are lower, due to the energy lost by friction.

Answer:

It is independent of the path of the body and depends only on the starting and ending points.

Explanation:

In Physics we define a conservative force as a force that is independent of the path of the body and depends only on the starting and ending points.

For conservative forces we can write;

KEi + PEi = KEf +PEf

where;

KEi= initial kinetic energy

PEi= initial potential energy

KEf= final kinetic energy

PEf= final potential energy

This equation is known as the principle conservation of mechanical energy . It applies only to conservative forces where friction is negligible. The term KE + PE  is also known as the total mechanical energy of the system.

Complete Question:

A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will

A. be behind the package.

B. be over the package.

C. be in front of the package.

D. depend on the speed of the plane when the package was released.

Answer:

B.

Explanation:

As no other horizontal forces are present, due to the horizontal movement and the vertical one are independent each other (as they are perpendicular), the plane and the package continue moving horizontally at the same speed, so when the package hits the ground (due to the action of gravity in the vertical direction only) the plane will be exactly over the package.

The package hits the ground in such a way that the plane will be exactly over the package.

The given problem is based on the effect of air resistance on the motion of object. The resistive force exerted on any object on account of dust, and other air particles is known as air resistance or simply as air drag.

In the given problem, as no other horizontal forces are present, due to the horizontal movement and the vertical one are independent each other (as they are perpendicular), the plane and the package continue moving horizontally at the same speed.

So when the package hits the ground (due to the action of gravity in the vertical direction only) the plane will be exactly over the package.

Thus, we can conclude that the package hits the ground in such a way that the plane will be exactly over the package.

Learn more about the air drag here:

https://brainly.com/question/14755232

Answer:

We are given:

initial velocity (u) = 20m/s

acceleration (a) = 4 m/s²

time (t) = 8 seconds

displacement (s) = s m

Solving for Displacement:

From the seconds equation of motion:

s = ut + 1/2 * at²

replacing the variables

s = 20(8) + 1/2 * (4)*(8)*(8)

s = 160 + 128

s = 288 m

Answer:

The Energy transferred to the engine by burning gasoline = 216.67 KJ

Explanation:

The parameters given are:

The efficiency of the car engine, E = 30% = 0.3

Mass, m = 1300 kg

Initial velocity, u = 0, since the car is from rest

The final velocity, v = 10 m/s

Since the car was moving, we calculate its kinetic energy.

kinetic energy = ((1/2) (m) (v^2)

((1/2) (1300 kg) (10 m/s^2)

= 65,000 j

The Energy, Q transferred to the engine by burning gasoline in this case

= potential energy / The efficiency of the car engine, E

Q = 65,000 j / 0.3

= 216,666.66 J

Converting Joule to kilojoule

where 1KJ = 1000j

216,666.66 J = 216.67 KJ

Answer: -9.8 m/s2

Explanation:

Answer:

The change in internal energy of the system is 2,054 J

Explanation:

The first law of thermodynamics relates the work and the transferred heat exchanged in a system through internal energy. This energy is neither created nor destroyed, it is only transformed.

Taking into account that the internal energy is the sum of all the energies of the particles that the system has, you have:

ΔU= Q + W

where U is the internal energy of the system (isolated), Q is the amount of heat contributed to the system and W is the work done by the system.

By convention, Q is positive if it goes from the environment to the system, or negative otherwise, and W is positive if it is carried out on the system and negative if it is carried out by the system.

In this case:

Replacing:

ΔU= 0.523 kJ + 1.531 kJ

Solving:

ΔU= 2.054 kJ = 2,054 J (being 1 kJ=1,000 J)

The change in internal energy of the system is 2,054 J

Answer:

Hello your question is incomplete attached below is the complete question

answer : 49 seconds

Explanation:

considering only Hysteresis loss

we have to calculate the Area affected/under the  Hysteresis loss

= volume * area

= 4 * ( 1.5 * 20 ) =  120 tesla. A/m

next we calculate the volume of the material

= mass of material / density

= 500 grams / 7.9 g/cm^3  = 6.33 * 10^-5 m^3

next we calculate the heat lost per cycle

= 6.33 * 10^-5 m^3 * 120 = 0.007596 joules

The total heat required to raise  temperature by 1°c

= Cp * ΔT * n

= 3R * n * ΔT  = 3(8.314) * 8.95 * 1  = 223.23 Joules

where n = number of moles = 500grams / 55.85 = 8.95moles

ΔT = 1

Therefore the time required to have to operate at a frequency of 60 cps

= Total heat required  / heat lost per cycle

=( 223.23 / 0.007596 ) 60 cps

= 489.796 seconds ≈  49 seconds

By cooling down the food, the thermal energy in the food molecules is reduced.

A refrigerator is an appliance that is commonly used in the home for the purpose of cooling down a substnace especially water and drinks.

Due to the fact that the molecules that compose matter are is in a state of constant random motion, the food molecules contain thermal energy. Hence, by cooling down the food, the thermal energy in the food molecules is reduced.

Learn more about thermal energy: https://brainly.com/question/11278589

Answer:

1>500gf

1>300gf

its answer

Answer:

The period of that same pendulum on the moon is 12.0 seconds.

Explanation:

To determine the period of that same pendulum on the moon,

First, we will determine the value of g (which is a measure of the strength of Earth's gravity) on the Moon. Let the value of g on the Moon be [tex]g_{M}[/tex].

From the question, the strength of earth’s gravity is only 1/6th of the normal value. The normal value of g is 9.8 m/s²

∴ [tex]g_{M}[/tex] = [tex]\frac{1}{6} \times 9.8 m/s^{2}[/tex]

[tex]g_{M}[/tex] = 1.63 m/s²

From the question, T=2π√L/g

[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]

We can write that,

[tex]T_{E} = 2\pi \sqrt{\frac{L}{g_{E} } }[/tex] .......... (1)

Where [tex]T_{E}[/tex] is the period of the pendulum on Earth and [tex]g_{E}[/tex] is the measure of the strength of Earth's gravity

and

[tex]T_{M} = 2\pi \sqrt{\frac{L}{g_{M} } }[/tex] .......... (2)

Where [tex]T_{M}[/tex] is the period of the pendulum on Moon and [tex]g_{M}[/tex] is the measure of the strength of Earth's gravity on the Moon.

Since we are to determine the period of the same pendulum on the moon, then, [tex]2\pi[/tex] and [tex]L[/tex] are constants.

Dividing equation (1) by (2), we get

[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]

From the question,

[tex]T_{E} = 4.9secs[/tex]

[tex]g_{E}[/tex] = 9.8 m/s²

[tex]g_{M}[/tex] = 1.63 m/s²

[tex]T_{M}[/tex] = ??

From,

[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]

[tex]\frac{4.9}{T_{M} } = \sqrt{\frac{1.63}{9.8} }[/tex]

[tex]\frac{4.9}{T_{M} } = 0.40783[/tex]

[tex]T_{M} =\frac{4.9}{0.40783 }[/tex]

[tex]T_{M} = 12.01 secs[/tex]

∴ [tex]T_{M} = 12.0secs[/tex]

Hence, the period of that same pendulum on the moon is 12.0 seconds.

Answer:

The period of that same pendulum on the moon is 12.0 s

Explanation:

Given;

period of a pendulum’s swinging, T=2π√L/g

the strength of earth’s gravity on moon, g₂ = ¹/₆(g₁)

period of pendulum on Earth, T₁ = 4.9 s

period of pendulum on moon, T₂ = ?

The length of the pendulum is constant, make it the subject of the formula;

[tex]T = 2\pi \sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\\\\(\frac{T}{2\pi} )^2 =\frac{L}{g}\\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\ L = \frac{gT^2}{4\pi^2}\\\\L_1 = L_2\\\\\frac{g_1T_1^2}{4\pi^2}= \frac{g_2T_2^2}{4\pi^2}\\\\g_1T_1^2 = g_2T_2^2\\\\T_2^2 = \frac{g_1T_1^2}{g_2} \\\\T_2 = \sqrt{\frac{g_1T_1^2}{g_2}}\\\\ T_2 = \sqrt{\frac{g_1T_1^2}{g_1/6}}\\\\ T_2 = \sqrt{\frac{6*g_1T_1^2}{g_1}}\\\\T_2 = \sqrt{6T_1^2}\\\\ T_2 = T_1\sqrt{6} \\\\T_2 = (4.9)\sqrt{6}\\\\ T_2 = 12.0 \ s[/tex]

Therefore, the period of that same pendulum on the moon is 12.0 s

Answer:

4.63rad/s

Explanation:

The centripetal force is expressed as;

F = mv²/r

M is the mass of the disk = 8kg

v is the linear speed = wr

r is the radius = 0.21m

Force = 36N

Write the force in term of the angular velocity

F = m(wr)²/r

F = mw²r²/r

F = mw²r

36 = 8w²(0.21)

36 = 1.68w²

w² = 36/1.68

w² = 21.43

w = √21.43

w = 4.63rad/s

Hence the angular speed is now 4.63rad/s

Answer:

-z

Explanation:

The force on a moving charge due to a magnetic field follows the right hand rule, so a positive charge, experiencing a magnetic deflection in the -y direction, while it moves in the direction of the x-axis, will do it  due to a magnetic field pointing in the +z direction.

As the electron has a negative charge, the magnetic field will point in the opposite direction, i.e., in the -z direction.

Answer:

Making the transition from young adulthood to middle adulthood can be difficult for some people. There are many changes which affect areas in a person’s biology, their psychology, their social life and their spiritual relationship. There are multiple stages of development which may affect an individual during middle adulthood which can be defined between either 30-65 years old or 40-65 years old.

Explanation:

Answer:

Option C.

Explanation:

To know which velocity-time graph best describes the motion of the ball, let us calculate the velocity of the ball and the time taken for the ball to get the ground. This can be obtained as follow:

1. Determination of the velocity.

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 100 m

Final velocity (v) =.?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 100)

v² = 0 + 1960

v² = 1960

Take the square root of both side.

v = √(1960)

v = 44.27 m/s

2. Determination of the time taken.

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 100 m

Time (t) =.?

h = ½gt²

100 = ½ × 9.8 × t²

100 = 4.9 × t²

Divide both side by 4.9

t² = 100 / 4.9

Take the square root of both side

t = √(100 / 4.9)

t = 4.52 s

From the above illustration,

Initial time (t1) = 0 s

Final time (t2) = 4.52 s

Initial velocity (u) = 0 m/s

Final velocity (v) = 44.27 m/s

Thus, we can see that as the time increase, the velocity also increase. Therefore, option C gives the correct answer to the question.

Answer:

46-D

47-C

48-F

49-A

50-B

I am not very sure I am right about those answers though.

Answer:You need to give more explanation sorry

Explanation:

Answer:

4.20 seconds

Explanation:

Supposing that silly goose weighs 69 pounds, we need to start on the math.

Simple maths, truly and really. 69/1=69, of course.

Therefore it will take 4.20 seconds for silly goose to hit the ground. if he is going to be a silly goose though, he can just go in the pond, instead of wasting his time.

Answer:

c. initial (x and y)

Explanation:

When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.

Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"

Thus, this method resolves the initial x and y velocities.