When two objects collide is the momentum before equal to the momentum after the collision? What happens to the individual carts momentum in terms of momentum before and after collision?

Answer:

yes

Explanation:

i hope this helps not sure im right

Answer:

The final velocity will be half of the initial velocity of the spacecraft.

Explanation:

Angular momentum is conserved for the circular force motion and central force motion.

Considering

L = MVR = Constant

Where

M = Mass of the object

V = Velocity of the object

r = radius of circle

We know that

V = [tex]\frac{1}{R}[/tex]

So,

[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{R_{1} }{R_{2} }[/tex]

As per the given data

[tex]R_{1}[/tex] = Initial Radius = 300 km

[tex]R_{2}[/tex] = Final Radius = 600 km

[tex]V_{1}[/tex] = Initial Velocity =

[tex]V_{2}[/tex] = Final Velocity =

Placing values in the formula

[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{300 km}{600 km }[/tex]

[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{1}{2}[/tex]

[tex]{V_{2}[/tex] = [tex]\frac{1}{2} V_{1}[/tex]

Hence, The final velocity will half of the initial velocity of the spacecraft.

Answer:

92127.2 j

Explanation:

Given that :

Mass of water = 290 g

Specific heat of water (C) = 4.18 j/g.C

Initial Temperature (t1) = 24°C

Final temperature (t2) = 100°C

Using the relation :

Q = mCdt

dt = change in temperature =( 100 - 24)° = 76°C

Q = quantity of heat ; C = specific heat capacity;

m = mass of substance

Q = 290 g * 4.18 j/g.C * 76°C

Q = 92127.2 j

Explanation:

Speed is the distance an object travel per unit of time

Answer:

The value is  [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]

Explanation:

From the question we are told that

   The  force is  [tex]F = 22.7 \ N[/tex]

    The value of room temperature is [tex]T = 298 \ K[/tex]

Generally the rate at which its entropy changes as it stretches is mathematically represented as

         [tex]\frac{\Delta S }{ L} = - \frac{F}{T}[/tex]

=>      [tex]\frac{\Delta S }{ L} = - \frac{21.5}{ 298 }[/tex]

=>      [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]

Answer:

Bus will take 4 seconds before it comes to rest.

Explanation:

Initial Velocity  of bus = u = 20 ms⁻¹

Final Velocity of bus = v = 0 ms⁻¹

Deceleration of bus = a = -5 ms⁻²

Time taken by bus before it comes to rest = t = ?

Using the equation formula

a = (v - u)/t

-5 = (0 - 20) / (t)

t = -20/-5

t = 4 sec

Therefore, Bus will take 4 seconds before it comes to rest.

Answer:

According to law of reflection i know in case of plane mirror the angle of reflection is equal to the angle of incidence. So,if you placed the eye on point A the hole can be seen.

Explanation:

Answer:

c

Explanation:

Answer:

What are the options?

Explanation:

Answer:

4.5 × 10^6 Nm

Explanation:

Given that a 52.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor covering. The heel on each shoe is circular and has a radius of 0.600 cm. If the woman balances on one heel, what pressure does she exert on the floor?

Let's first calculate the area covered by the heel shoe by using the area of a circle.

Convert cm to m

0.6 / 100 = 0.006

A = πr^2

A = 22/7 × 0.006^2

A = 1.131 × 10^-4 m^2

The force exerted by the woman = mg

Force exerted = 52 × 9.8

Force exerted = 509.6 N

Pressure = force / Area

Pressure = 509.6 / 1.131 × 10^-4

Pressure = 4505853.3 N/m

Therefore, she exerts pressure of 4.5 × 10^6 Nm approximately on the floor.

Answer:

1.25s

1) 90.9m

Explanation:

From the question, we are given the following;

Speed v = 30m/s

Maximum height H = 45m

Required;

Time

Using the equation of motion S = ut + 1/2gt²

Substitute;

45= 30t + 1/2(9.8)t²

45 = 30t + 4.9t²

4.9t² + 30t - 45 = 0

Factorize;

t = -30 ±√30²-4(-45)(4.9)/2(4.9)

t = -30 ±√900+882/9.8

t = -30 ±42.21/9.8

t = -30 + 42.21/9.8

t = 12.21/9.8

t = 1.25secs

Hence the ball spent 1.25secs in air.

i) To get the horizontal distance, we will use the formula;

R = U√2H/g

R = 30√2(45)/9.8

R = 30√90/9.8

R = 30√9.18

R = 30(3.03)

R = 90.9m

Hence the horizontal distance is 90.9m

Complete Question

the maximum force a pilot can stand is about seven times his weight. what is the minimum radius of curvature that a jet plane's pilot, pulling out of a vertical dive, can tolerate at a speed of 250m/s?

Answer:

The value is    [tex]r = \frac{250^2 }{6 * 9.8 }[/tex]

Explanation:

From the question we are told that

 The  weight of the pilot is   [tex]W = mg[/tex]

 The maximum force a pilot can withstand is  [tex]F_{max} = 7 W = 7 (mg)[/tex]

 The speed is  [tex]v = 250 \ m/s[/tex]  

Generally the centripetal force acting on the pilot is equal to the net force acting on the pilot i.e

      [tex]F_c = F_{max} - mg[/tex]

Here N  is the normal force acting on the pilot

Now

      [tex]F_c = \frac{m v^2 }{r}[/tex]

So

      [tex]\frac{m v^2 }{r} = 7(mg) - mg[/tex]  

=>  [tex]r = \frac{v^2 }{6g}[/tex]

=>  [tex]r = \frac{250^2 }{6 * 9.8 }[/tex]

=>  [tex]r = 1063 \ m[/tex]

Explanation:

Force, F = 60 N

Angle, [tex]\theta=30^{\circ}[/tex]

We need to find horizontal and vertical component of the force.

Horizontal component,

[tex]F_x=F\cos\theta\\\\=60\times \cos(30)\\\\=51.96\ N[/tex]

Vertical component,

[tex]F_=F\sin\theta\\\\=60\times \sin(30)\\\\=30\ N[/tex]

So, the horizontal and vertical component are 51.96 N and 30 N respectively.

Answer:

The correct option is  c

Explanation:

From the question we are told that

    The ratio of the noise to human hearing threshold is  [tex]I = 150 000[/tex]

Generally from the equation given we have that

       dB =  10 log I

So

      dB =  10 log 150000  

The expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000. Option C is correct

Given the equation for calculating the intensity of sound which is measured in decibel expressed as:

dB=10 log I

where;

I is the ratio of the sound to the human hearing threshold

Given that noise is 150,000 times greater than the human hearing threshold.

Substitute I = 150,000 into the expression above;

dB =  10 log 150,000

Hence the expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000

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Answer:

stopping potential is the negative potential applied to the circuit to stop the moving electrons so as to stop the flow of current

for high current high negative potential is applied

Answer:

120 km/hr

Explanation:

Let D be the distance between the rocket and the camera as the rocket is moving upwards. Let d be the distance the rocket moves and L be the distance between the camera and the base of the rocket = 4 km.

Now, at any instant, D² = d² + L²

= d² + 4²

= d² + 16 since the three distances form a right-angled triangle with the distance between the rocket and the camera as the rocket is moving upwards as the hypotenuse side.

differentiating the expression to find the rate of change of D with respect to time, dD/dt ,we have

d(D²)/dt = d(d² + 16)/dt

2DdD/dt = 2d[d(d)/dt]

dD/dt = 2d[d(d)/dt] ÷ 2D

Now d(d)/dt = vertical speed of rocket = 200 km/hr

dD/dt = 200d/D  [D = √(d² + 16)]

dD/dt = 200d/[√d² + 16]

Now substituting d = 3 km, the distance the rocket has risen into the equation, we have

dD/dt = 200(3)/[√(3² + 16)]

dD/dt = 600/[√(9 + 16)]

dD/dt = 600/√25

dD/dt = 600/5

dD/dt = 120 km/hr

So,  the speed at which the distance from the camera to the rocket changing when the rocket has risen 3 km is 120 km/hr.

Answer:

yes it also does have but not in the exact form but it does 30 percent of respiration to produce rain

Answer:

[tex]a = 2.56\ m/s^2[/tex]

Explanation:

Mechanical Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The object has a mass of m=430 g = 0.43 Kg and is pushed by a net force of F=1.1 N. To calculate the acceleration, we solve for a:

[tex]\displaystyle a=\frac{F}{m}[/tex]

[tex]\displaystyle a=\frac{1.1}{0.43}[/tex]

[tex]\mathbf{a = 2.56\ m/s^2}[/tex]

This question is incomplete, the complete question is;

A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass ;

a) F_throw = 8.083 N

b) F_throw = 9.181 N

c) F_throw = 2.284 N

d) F_throw = 16.014 N

e) None of these is correct

Answer:

the quarterback applied a force of 8.083 N to the ball during the pass

so Option a) F_throw = 8.083 N is the correct answer

Explanation:

Given that;

m = 0.408 kg

d = 1.909 m

u = 0 { from rest}

t = 0.439 s

Now using Kinetic equation

d = ut + 1/2 at²

we substitute

1.909 = (0 × 0.439) + 1/2 a(0.439)²

1.909 = 0 + 0.09636a

1.909 = 0.09636a

a = 1.909 / 0.09636

a = 19.8111 m/s²

Now force applied will be;

F = ma

we substitute

F = 0.408 ×  19.8111

F = 8.0828 ≈ 8.083 N

Therefore the quarterback applied a force of 8.083 N to the ball during the pass

so Option a) F_throw = 8.083 N is the correct answer

1 spring k=100

2 springs k =200

f=kx

then k=f/x

200=150/x

x=200/150

x=4/3

Explanation:

need thanks ..................

The total extension caused by the 150N weight is 1.5 m

From Hooke's law,

Force (F) = spring constant (K) × extention (e)

F = Ke

150 = 100 × e

Divide both side by 100

e = 150 / 100

e = 1.5 m

Therefore, the total extention caused by the weight is 1.5 m

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Answer:

The answer is true, as gravity affects everything.

Answer:

Explanation: about 2 Mph when it falls and sorry if I’m wrong

Answer:

Energy is applied on the charge to do work.

Explanation:

Work is only be done when one charge moves against the electric field of anther charge that require huge amount of energy because both have same charge and the force of repulsion occurs between them. Work is defined as the product of force and displacement so from this equation we can conclude that work is done when a force is applied on an object and it moves in the direction in which force is applied. If the force or energy is removed from the charge which is present in the electric field of another charge so it moves away from that charge and the work is also be done..

Answer:

use hot water

Explanation:

hot water helps dissolve things faster

Answer:

The surface charge density on planes A and B respectively is

     [tex]\sigma__{A}} } = 2\sigma[/tex]

and

     [tex]\sigma__{B}} = \sigma[/tex]

Explanation:

From the question we are told that

     The  electric field in region to the left of  A  is  [tex]E_i = \frac{3 \sigma}{2 \epsilon_o}[/tex]

      The  direction of the electric field is  left

      The  electric field in the region to the right of  B  is  [tex]E_f = \frac{3 \sigma}{2 \epsilon_o}[/tex]

      The  direction of the electric field is  right

      The  electric field in the region between the two planes is  [tex]E_m = \frac{\sigma }{2 \epsilon_o }[/tex]

       The  direction of the electric field is  right

Let the surface charge density on planes A and B be represented as [tex]\sigma__{A}} \ \ and \ \ \sigma__{B}} \ \ \ respectively[/tex]

 From the question we see that

         [tex]E_i = E_f[/tex]

Generally the electric to the right and to the left is due to the combined electric field generated by  plane  A and  B  so

         [tex]E_i = E_f = \frac{3\sigma }{2\epsilon} = \frac{\sigma_A }{ 2 \epsilon_o } + \frac{\sigma_B }{ 2 \epsilon_o }[/tex]

=>      [tex]\sigma__{A}} + \sigma__{B}} = 3 \sigma -- -(1)[/tex]

Generally the electric field at the middle of the plane A and  B  is due to the diffencence in electric field generated by plane A and  B      

i.e

          [tex]\frac{\sigma }{2 \epsilon_o } = \frac{\sigma_A }{ 2 \epsilon_o } - \frac{\sigma_B }{ 2 \epsilon_o }[/tex]

=>      [tex]\sigma__{A}} - \sigma__{B}} = \sigma[/tex]

=>      [tex]\sigma__{A}} } = \sigma + \sigma__{B}[/tex]

From equation 1

        [tex]\sigma + \sigma__{B}}+ \sigma__{B}} = 3 \sigma[/tex]

=>    [tex]\sigma__{B}} = \sigma[/tex]

So

      [tex]\sigma__{A}} } = \sigma + \sigma[/tex]

=>   [tex]\sigma__{A}} } = 2\sigma[/tex]

Answer:

The current moving through tha circuit changes direction.

Explanation:

hope this helps

The electromagnet is activated. It causes a buzzer to go off. The hammer strikes the bars which creates sound. When you let go, it will stop making a sound by the circuit stop flowing.

"Electromagnet, device consisting of a core of magnetic material surrounded by a coil through which an electric current is passed to magnetize the core. An electromagnet is used wherever controllable magnets are required, as in contrivances in which the magnetic flux is to be varied, reversed, or switched on and off."

"A circuit is a complete circular path that electricity flows through. A simple circuit consists of a current source, conductors and a load. The term circuit can be used in a general sense to refer to any fixed path that electricity, data or a signal can travel through."

Know more about electromagnet here

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Answer:

1. Brown dwarf.

2. Red giant.

3. Black dwarf.

4. White dwarf.

5. Black hole.

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

Basically, stars are typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).

Some of the examples of stars are; Canopus, Sun (closest to the Earth), Betelgeus, Antares, Vega etc.

1. Brown dwarf: does not undergo hydrogen fusion.

2. Red giant: has an outer shell of burning hydrogen and a core of helium.

3. Black dwarf: a white dwarf that has lost all its energy.

4. White dwarf: has a core of carbon that is not massive enough to fuse. A white dwarf also known as degenerate dwarf is a star or an electron degenerate matter and is as massive as the Sun but only about as large in size as planet Earth. Also, the maximum mass of a white dwarf is about 1.4 times the mass of the Sun.

5. Black hole: formed from the collapse of a white dwarf beyond 1.5 solar masses.

Answer:

other person is correct!!!

Explanation:

got 100 on test