What force is necessary to accelerate a 2500 kg care from rest to 20 m/s over 10 seconds?

A few years ago, an X-ray telescope detected a source called Cygnus X-3, whose intensity changed with a period of 4.8 hours. Thi

alina1380 [7]

Answer:

Explanation:

The star is revolving the black hole like earth revolves around the sun .so time period of rotation T is given by the following relation

T² = $\frac{4\pi^2\times R^3}{GM }$ , R is distance between black hole and star , M is mass of black hole

Given T = 4.8 hours

4.8² = $\frac{4\pi^2\times R^3}{GM }$

Using the same equation for earth sun system

24² = $\frac{4\pi^2\times (50R)^3}{GM_s }$ , Ms is mass of the sun and 50R is distance between the sun and the earth .

Dividing the equation

$(\frac{4.8}{24})^2$ = $\frac{M_s}{M}\times\frac{1^3}{50^3}$

$\frac{M}{M_s}$ = 2x 10⁻⁴

7 0

3 years ago

PLEASE HELP The image characteristics are...

VLD [36.1K]

According to the question, the object is placed at 2F

The ray diagram is shown in the figure attached.

According to the figure:

Object AB is at 2F₁

First, we draw a ray parallel to principal axis.

So, it passes through focus after refraction.

We draw another ray which passes through optical center.

So, the ray will go through without any deviation.

Where both refracted rays meet is point A' and the image formed is A'B'

This image is formed at 2F₂

We can say that:

Image is real.
Image is inverted.
Image is exactly the same size as the object.

4 0

3 years ago

what happens to a circuits resistance, voltage, and current when you increase the diameter of the wire in the circuit?

zhannawk [14.2K]

Answer:

Option D.

Resistance (R) decreases

Voltage (V) is constant

Current (I) increases

Explanation:

We'll begin by writing an equation relating resistance and diameter of a wire. This is given below:

R = ρL/A ......... (1)

A = πr² (since the wire is circular in shape)

r = d/2

A = πr² = π(d/2)²

A = πd²/4

Substitute the value of A into equation 1

R = ρL/A

R = ρL ÷ A

R = ρL ÷ πd²/4

R = ρL × 4/πd²

R = 4ρL /πd²

Where:

R is the resistance of the wire.

ρ is the resistivity of the wire.

L is the length of the wire.

A is the cross sectional area of the wire.

r is the radius.

d is the diameter of the wire

From equation (1) above, we can say that the resistance (R) is inversely proportional to the square of the diameter of the wire. This implies that an increase in the diameter of the wire will result in a decrease of the resistance. Also, a decrease in the diameter of the wire will result in an increase in the resistance of the wire.

1. Since the diameter of the wire is increase, therefore, the resistance of the wire will decrease.

2. From ohm's law,

V = IR

Divide both side by I

R = V/I

Where:

R is the resistance

V is the voltage

I is the current

From the above equation, the resistance (R) is directly proportional to the voltage (V) and inversely proportional to the current (I).

If we keep the voltage constant, this means that an increase in the resistance will lead to a decrease in the current. Also, a decrease in the resistance will lead to an increase in the current.

Since the resistance of the wire decrease, the current will increase.

From the illustrations made above, an increase in the diameter of the wire will lead to:

1. Decrease in resistance.

2. Voltage is constant.

3. Increase in current.

5 0

3 years ago

Think of a body which is at rest but not in equilibrium

REY [17]

Answer:

I think the sun is the right answer

8 0

3 years ago

A shuffleboard player pushes a 0.25 kg puck, initially at rest, such that a constant horizontal force of 6 N acts on it through

AveGali [126]

Answer:

(a) 3 J and 4.899 m/s

(b) -3 J.

(c) It will take four times as much as the work done to stop it when the final speed is increased twice.

Explanation:

(a) Work done by the shuffleboard player = Force × distance

W = F×d.................................. Equation 1

Where W = work done, F = Force, d = distance.

Kinetic Energy (Ek) of the = 1/2mv²...................... Equation 2

Where m = mass of the puck, v = velocity of the puck.

Note: Work done by the shuffleboard player = Kinetic Energy of the puck when the force is removed, assuming no energy is lost to friction.

Therefore,

Ek = 1/2mv² = F×d

Ek = F×d ............................................. Equation 3

Given: F = 6 N, d = 0.5 m.

Substituting these values into equation 3

Ek = 6×0.5

Ek = 3 J.

Thus the kinetic Energy = 3 J.

Also,

Ek = 1/2mv²

making v the subject of the equation

v = √(2Ek/m)....................... Equation 4

Given: m = 0.25, Ek = 3 J

Substituting into equation 4

v = √(2×3/0.25)

v = √24

v = 4.899 m/s

Thus the speed of the puck = 4.899 m/s

(b) The work required to bring the puck to rest = -(the Kinetic Energy of the puck.)

Wₙ = 1/2mv²

Where Wₙ = work required to bring the puck to rest.

Where m = 0.25 kg, v = 4.899 m/s²

Wₙ = -1/2(0.25)(4.899)²

Wₙ = -1/2(0.25)(24)

Wₙ = -0.25(12)

Wₙ = -3 J

Thus the work required to bring the puck to rest = 3 J.

(c) Assuming the puck has twice the final speed

Work required to stop it.

Wₓ = 1/2mv²

Where Wₓ = work required to stop the puck when it has twice the final speed.

m = 0.25 kg, v = 9.798 m/s ( twice the final speed)

Wₓ = 1/2(0.25)(9.798)²

Wₓ = 1/2(0.25)(96)

Wₓ = 12 J.

Thus the puck will take four times as much as the work done to stop it when the final speed is increased twice.

3 0

3 years ago