Which is the last stage in the life cycle of an average star? (1 point)

A.white drawf

B.super red giant

C. planetary nebula

D.neutron star

The force must be applied for motor to keep the package moving is 50 N.

Force is the action of push or pull the object in order to make it move or stop.

Power is related to force and velocity as

P = F x v

150 W = F x 3 m/s

F  =50 N

Thus, the force must be applied for motor to keep the package moving is 50 N.

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Answer:

Rocket 4

Explanation:

since it has the smallest mass it is going to the bigger acceleration.

Answer:

C

Explanation:

In theory, rocket 4 has the least mass and highest force output in comparison.

Answer:

A - energy is absorbed in the reaction

Explanation:

Answer:

b

Explanation:

The formula for work done (W) is:

W = Force x Distance moved in the direction of force

This means that the object needs to move in the direction of the force, otherwise no work is done in that direction.

Hope this helps!

Answer:

11 is d. 12 is c. 13 is b. 14 is b. 15 is b.

Explanation:

The magnitude of that other charge will be 9×10⁻⁵ C. The force on the charge is inverse of the distance.

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The magnitude of that other charge is found as;

[tex]\rm F = \frac{Kq_1q_2}{r^2} \\\\ \rm 9 = \frac{9 \times 10^9 \times 1\times 10^{-4}\times q_2}{(3)^2} \\\\ q_2=9\times 10^{-5} \ C[/tex]

Hence, the magnitude of that other charge will be 9×10⁻⁵ C.

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Answer:

A) 75m

Explanation:

distance = speed x time taken

Here given:

speed: 15 m/s

time: 5 second

So, distance covered:

15 x 5

75 m

Distance

Answer:

To express this concept mathematically, the work W is equal to the force f times the distanced, or W = fd. If the force is being exerted at an angle θ to the displacement, the work done is W = fd cos θ.

The SI unit of work is the joule (J). It is defined as the work done by a force of one newton through a distance of one meter.

Explanation:

Work is done when a force is applied to an object through a distance. This means that when a force is applied to an object through a distance, the object's total energy will be affected.

Answer:

300 J

Explanation:

GIVEN:

Force (F) = 10 N

Displacement (s) = 30 m

or displacement (d) = 30m

Work done (W) = ?

Therefore,

W = F×s

or W= F × d

= 10×30

= 300 J

Hope it helps

The direction of force refers to the path along which the force acts. The plane of application of force, on the other hand, is the surface or area where the force is applied

The direction of force refers to the path along which the force acts. It is the straight line in which the force is being applied. For example, if you push a box on the ground, the direction of force is the line of action of the force vector that points in the direction of the push.

The plane of application of force is the surface or area where the force is applied.

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Answer:

23.24 m/s

Explanation:

2.1 s  *  (-7.6 m/s^2) = -  15.96 m/s   from original speed

39.2 - 15.96 = 23.24 m/s

Answer:

D

Explanation:

Because we have two reactants and product

Answer:

.Also we can change the frequency by decreasing wavelength , or increasing velocity

Explanation:

Answer:

4 m/s

Explanation:

v=u+ at  and  v^2=u^2+2as.

It is given that  v=0,t=5  s    and  s=10  m.

⇒0=u+5a  and  0=u^2+20a.

⇒a=−u/5  and  a=−u^2/20.

⇒u/5 = u^2/20⇒u=4  m/s.

⇒  The initial velocity is  4  m/s.

Answer:

To be better at something

Explanation:

I have Accelerated at cooking

Answer:

Acceleration means to speed up

Explanation:

Answer:

A) P = W / t = 6000 / 60 = 100 J/s == 100 watts

B) P = I V = .05 * 240 = 12 watts

C)  Bulb B) uses much less power

Answer:

3+7=          7x3=                                21➗7=          

Explanation:

3+7= 10              7 8 9 10

7x3= 21                  7 14 21

21➗7=3        21 14 7

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Answer:

See below

Explanation:

Vertical position = 45 +  20 sin (30) t  - 4.9 t^2

 when it hits ground this = 0

               0 = -4.9t^2 + 20 sin (30 ) t + 45

                0 = -4.9t^2 + 10 t +45 = 0     solve for t =4.22 sec

  max height is at  t= - b/2a = 10/9.8 =1.02

     use this value of 't' in the equation to calculate max height = 50.1 m

      it has  4.22 - 1.02 to free fall = 3.2 seconds free fall

           v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL

      it will also still have horizontal velocity =  20 cos 30 = 17.32 m/s

        total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s

Horizontal range = 20 cos 30  * t  =  20 *  cos 30  * 4.22 = 73.1 m

Answer:

Length of steel = 5.5 m

Increase in length = 0.0012 m

α of steel = 12 x 10-6/K

To find:

Temperature change needed ?

Calculation:

Let the required temperature change be ∆T. (in the photo)

So, temperature change of 18.1 Kelvin is required

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Answer:

doppler

Explanation:

Answer:

0.12m/s

Explanation:

v=λf

Given that, λ = 12cm = 0.12m

T = 1second

(A period T is the time required for one complete cycle of vibration to pass a given point)

frequency 'f' is unknown but we can get frequency from f = 1/T = 1/1 = 1Hz

therefore, v= 0.12 × 1 = 0.12m/s

The average velocity of the boat given the data from the question is 6.32 Km/h

Velocity is defined as the rate of change of displacement with time. Mathematically, it is expressed as

Velocity = displacement / time

D² = d₁² + d₂²

D² = 3² + 1²

D² = 9 + 1

D² = 10

Take the square root of both sides

D = √10

D = 3.16 Km

Average velocity = resultant displacement / time

Average velocity = 3.16 / 0.5

Average velocity = 6.32 Km/h

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Answer:

a.one atom of fluorine and two atoms of oxygen

Explanation:

hope it help

Answer:

True

Explanation:

If it is at right conditions and correct time (night time most likely) they will all be visible

The total momentum before collision is

14.0 kg•m/s + (-9.00 kg•m/s) = 5.00 kg•m/s

If v is the velocity of ball B after colliding with ball A, the total momentum after collision is

(2.00 kg) (-5.00 m/s) + (3.00 kg) v

Momentum is conserved, so these quantities are equal. Solve for v :

5.00 kg•m/s = (2.00 kg) (-5.0 m/s) + (3.00 kg) v

5.00 kg•m/s = -10.0 kg•m/s + (3.00 kg) v

15.0 kg•m/s = (3.00 kg) v

v = (15.0 kg•m/s) / (3.00 kg) = 5.00 m/s

Answer:

Here is your answer

Explanation:

Given:

Given:Density of sea water (d) = 1020 kgm^{3}3

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,P = h × d × g

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,P = h × d × g\therefore∴ h = P/dg

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,P = h × d × g\therefore∴ h = P/dg\begin{gathered} = \frac{10200000}{1020 \times 10} \\ \\ = 1000 \: m \\ \\ = 1 \: km.\end{gathered}=1020×1010200000=1000m=1km.

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,P = h × d × g\therefore∴ h = P/dg\begin{gathered} = \frac{10200000}{1020 \times 10} \\ \\ = 1000 \: m \\ \\ = 1 \: km.\end{gathered}=1020×1010200000=1000m=1km.Hence, depth of sea is 1 km