The weight of a 0.5 kg object on the surface of Planet X is 20 N. If the radius of the planet is 4 X 106 m, what is its mass?

Answer:

77000 J

Explanation:

Formula for the heat energy is;

Q = m•c•Δt

We are given;

mass; m = 2 kg

Change in temperature; Δt = 110 - 10 = 100 °C

From online values, specific heat capacity of copper is; c = 385 J/kg.°C

Thus;

Q = 2 × 100 × 385

Q = 77000 J

Answer:

heat = 20 Kcal

Explanation:

Answer:

zero

Explanation:

no distance has moved while holding the sheet...so no distance means no workdone..no workdone means no power...

Answer:

40 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Work done is simply defined as the product of force and distance moved in the direction of the force. Mathematically, we can express the Workdone as:

Workdone = force × distance

Wd = F × s

With the above formula, we can obtain the workdone as follow:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Wd = F × s

Wd = 10 × 4

Wd = 40 J

Thus, 40 J of work was done.

Answer:

s = 2.79 m

Explanation:

Given that,

A ball have a speed of 7.4 m/s just before it hits the ground.

Initial velocity of the ball was 0 (at rest)

We need to find the height from where the ball was dropped. It means we need to find the distance covered by it. Let it is h.

Using the third equation of motion to find it as follows :

[tex]v^2-u^2=2as\\\\s=\dfrac{v^2}{2g}\\\\s=\dfrac{(7.4)^2}{2\times 9.8}\\\\s=2.79\ m[/tex]

So, the ball is dropped from a height of 2.79 m.

The height from which you would have to drop a ball so that it would have a speed of 7.4 m/s just before it hits the ground is 2.79m

According to the third equation of motion;

[tex]v^2=u^2+2as[/tex]

where;

v is the final velocity = 7.4m/s

Initial velocity = 0m/s

s is the distance

a = g= 9.8m/s²

Substitute the values into the formula to have:
[tex]7.4^2=0^2+2(9.8)s\\54.76 + 19.6s\\s=\frac{54.76}{19.6}\\ s=2.79m[/tex]

Hence the height from which you would have to drop a ball so that it would have a speed of 7.4 m/s just before it hits the ground is 2.79m

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Answer:

a refractor telescope

Explanation:

Answer:

a. refractor telescope

The question is missing some parts. Here is the complete question.

A suspension bridge cable is connected to its anchor at a 20° angle. Find the vertical and horizontal component of the force on the anchor by the cable.

Answer: [tex]F_{x}=[/tex] 14095.4 N

              [tex]F_{y}=[/tex] 5130.3 N

Explanation: The force applied to the anchor is not perpendicular to the horizontal plane. So, it can be decomposed into 2 components: a vertical component, which is on the y-axis, and a horizontal component, which is on the x-axis.

The force and its components forms a right triangle, so we can calculate the components by using trigonometric relations:

Horizontal

[tex]cos(20)=\frac{F_{x}}{F}[/tex]

[tex]F_{x}=F.cos(20)[/tex]

[tex]F_{x}=15,000(0.9397)[/tex]

[tex]F_{x}=[/tex] 14,095.4 N

Vertical

[tex]sin(20)=\frac{F_{y}}{F}[/tex]

[tex]F_{y}=Fsin(20)[/tex]

[tex]F_{y}=[/tex] 15,000(0.3420)

[tex]F_{y}=[/tex] 5,130.3 N

The vertical and horizontal components of force on the anchor by the cable are 5130.3 N and 14095.4 N, respectively.

Answer:

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 12 × 24

We have the final answer as

Hope this helps you

Answer:

2.9m/s and 140N

Explanation:

For speed convert rev/s to m/s

(Rev/8s)=(2π/8)(5m) = 3.9m/s

For Centripetal Force

Fc = (ac)(m) = (v^2/r)(m) = [(3.9m/s)^2/(5.0m)][45kg] = 140N

Answers go to two significant figures.

Answer:

F = 133.33[N]

Explanation:

This problem can be solved by the principle of momentum conservation, which tells us that momentum is preserved before and after the bounce of the ball on the wall.

In such a way that the movement towards the wall we will take it with a positive sign, and the force of the rebound to the left as negative. The movement to the left will be taken as a negative sign.

[tex]m_{1}*v_{1}-F*t=-m_{1}*v_{2}[/tex]

where:

m₁ = mass of the object = 1.5 [kg]

v₁ = velocity of the ball before hitting the wall = 2.5 [m/s]

F = average force [N]

t = time contact = 0.045 [s]

v₂ = velocity of the ball after hitting the wall = 1.5 [m/s]

Now replacing:

[tex](1.5*2.5)-F*0.045=-(1.5*1.5)\\3.75+2.25=F*0.045\\F=6/0.045\\F=133.33[N][/tex]

Answer:

0J

Explanation:

No work is being done on the wall by the man pushing on it.

 Given parameters:

Mass of man  = 75kg

Mass of wall  = 500000kg

Time  = 250s

Unknown:

Work done  = ?

Solution:

Work done is the force applied on a body that moves it along a particular path.

For work to be done, distance must be move or displacement must occur.

Since the wall is not moving the distance is 0;

    Work done  = Force x distance

Since distance is 0m, work done is 0J

The work done on the wall by the man is 0 J.

To calculate the amount of work done by the man, we use the formula below.

Formula:

Where:

from the question,

Given:

Substitute these values into equation 1

Hence, the work done on the wall by the man is 0 J.

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Answer:

7. d. England during the Age of Enlightenment

8. a. State governments do not usually act together

Explanation:

The Age of Enlightenment occurred in the 18th century and at this time England was a Constitutional monarchy with a monarch and parliament ruling the country with different powers.

The third and fourth methods of amending the Constitution will be much harder to use because state governments simply do not usually act together as they most times have ideological differences. The sheer number of states it would take to unite for these methods to be used makes this impractical.

Answer:

41.02m

Explanation:

Given parameters:

Initial velocity = 0m/s

Final velocity  = 96km/hr

Time taken  = 3.07s

Unknown:

Distance traveled by the time the final speed was achieved = ?

Solution:

To solve this problem, we first find the acceleration of the car;

     Acceleration  = [tex]\frac{v - u }{t}[/tex]

v is the final velocity

u is the initial velocity

t is the time taken

  Now convert the the final velocity to m/s;

          96km/hr to m/s;

               1 km/hr  = 0.278m/s

               96km/hr  = 96 x 0.278 = 26.7m/s

Now;

 Acceleration  = [tex]\frac{26.7 - 0}{3.07}[/tex]   = 8.69m/s²

So;

   v²  = u²  + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

             26.7²  = 0²  + 2 x 8.69 x s

             712.89  = 17.38s

                  s  = 41.02m

Answer:

t = 3/8 seconds

Explanation:

h=-16t^2 - 10t+6

h= 0 when it hits the ground

0=-16t^2 - 10t+6

factor out a -2

0= -2(8t^2 +5t -3)

divide by -2

0 = (8t^2 +5t -3)

factor

0=(8t-3) (t+1)

using the zero product property

8t-3 = 0    t+1 =0

8t = 3         t= -1

t = 3/8     t= -1

t cannot be negative  ( no negative time)

t = 3/8 seconds

Explanation:

Given parameters:

Height  = 5m

Mass of Sue  = 50kg

Unknown:

Magnitude of Sue's weight  = ?

Work done by Sue = ?

Solution:

Weight is the vertical force exerted by a body in the presence of gravity.

Mathematically;

        W = mg

m is the mass

g is the acceleration due to gravity  = 9.8m/s²

  Weight  = 50 x 9.8  = 490N

Work done  = Force x distance = weight x height

 Work done  = 490 x 5 = 2450J

Answer:

[tex]D= -0.42km[/tex]

Explanation:

From the question we are told that

Drifting right with speed 5.0m/s

The motor runs for 6.0 seconds

Leftward acceleration of magnitude 4.0 m/s squared

Generally the equation [tex]V=ut+1/2at^2[/tex] can be used here

[tex]V=ut+1/2at^2[/tex]

Mathematically solving with the newton equation above we have that

   [tex]D=5*6 + \frac{1}{2} (-4)*6^2[/tex]

   [tex]D=30-72[/tex]

   [tex]D=-42m[/tex] [tex]or -0.42km[/tex]

Therefore having this the Displacement is [tex]D= -0.42km[/tex] leftward

Answer:

A. 1600 kilojoules

Explanation:

Looking at the graph, at 4 meters, we can guess the energy transferred would be around 1500 kilojoules. 1600 kilojoules is the closest to that value, so A would be the best choice.

Answer:

one could freeze and the second would thaw

Explanation:

sorry if its wrong

The two examples of how land can have a dramatic change in temperature

is during freezing and thawing.

Cold temperatures which is common during the winter period is

characterized by the formation of snow and freezing of smaller water

bodies.

There may be a short phase in which there is relative sunlight which melts

the frozen substances thereby forming liquids . This is usually as result of

the temperature being on the increase in the atmosphere.

Read more on https://brainly.com/question/25671319

Answer:

The answer would be 11.6 g/m^3

Explanation:

Knowing that vapor's density at 100% humidity is 20g/m^3 we can calculate the answer using the following math:

0.58 * 20 = 11.6 grams per cubic meter

the units can also be written as g/m^3

The density of water vapor is 11.6 grams/cubic meter.

The quantity of atmospheric moisture that is present compared to the amount that would be there if the air were saturated is known as relative humidity, and it is stated as a percentage. Relative humidity depends on both moisture content and temperature because the latter quantity is temperature-dependent. The related temperature and dew point for the specified hour are used to calculate relative humidity. it is expresses as %.

Now given that the maximum water vapor in air at this temperature is 20 grams/cubic meter at 22°C, that is, density of vapor at 100% humidity is 20g/[tex]m^3[/tex]  at 22°C.

And, The device showing that at 22 degrees Celsius the relative humidity is 58%

So, water vapor density at that moment is = 20g/[tex]m^3[/tex] × 58%

= 11.6 g/[tex]m^3[/tex] .

= 11.6 grams/cubic meter.

Hence, according to the measurement of device, the water vapor density is 11.6 grams/cubic meter.

Learn more about relative humidity here:

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Answer:

c. joules per gram degree Celsius

Explanation:

edg 2021

Answer:

3743.489 kg

Explanation:

F_g = 591 N

G = 6.674x10^-11 constant of gravity

m_1 = 95 kg

m_2 = unknown

r = 4990*1000 =

F_g = G[(m_1*m_2)/r^2]

591 N = 6.674x10^-11[(95*m_2)/4990^2]

8.855 = [(95*m_2)/4990^2]

355631.472 = 95*m_2

m_2 = 3743.489 kg

The  mass of the planet, which has a radius of 4990 km, is  1.81×10²³kg.

Newton's law of universal gravitation states that

The force of attraction between any two bodies is inversely proportional to the square of the distance between them and directly proportional to the product of their masses.

Given parameters:

Mass of the object: m = 95 kg

Radius of the planet: r = 4990 km = 4990 × 1000 m.

Weight f the object: F= 591 N

We know that: universal gravitational constant: G = 6.674x10^-11 SI unit.

We have to find: mass of the planet: M = ?

Now, F= GMm/r^2

591 N = 6.674x10^-11[(95×M)/(4990×1000)^2]

⇒ M =1.81×10²³kg

Hence, mass of the planet is  1.81×10²³kg.

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Answer: = 2,496 N

Explanation:

Force is the push or pull on an object and it is calculated as:

= Mass * Acceleration

= 1,248 * 2.0

= 2,496 N

Answer: 85J

Explanation:

From the question, we are informed that when spring was wound, 100J of work was done but 15J escaped to the surrounding as heat.

Therefore, the change in internal energy of the spring will be calculated as:

ΔU = q + w

where, q = -15J

w = 100J

ΔU = -15J + 100J

= 85J

Answer: 432m

Explanation:

Convert 3 min to seconds

1 min = 60 sec

3 min = 180 sec

Multiply the speed times time to get distance.

2.4 x 180 = 432m

Answer:

The minimum uncertainty in its position is 1.85 x 10⁻³² m.

Explanation:

Given;

mass of the ball, m = 47.5 g = 0.0475 kg

speed of the ball, v = 30 m/s

measuring accuracy of the speed, = 0.2% = 0.002

The uncertainty in measurement of momentum;

ΔP = mΔv

ΔP = (0.0475)(30 x 0.002)

ΔP = 2.85 x 10⁻³ kgm/s

The uncertainty in position is calculated as;

[tex]\delta x \geq \frac{h}{4\pi (\delta P)}[/tex]

where;

h is Planck's constant

[tex]\delta x \geq \frac{6.626 \ \times \ 10^{-34}}{4\pi (2.85 \ \times \ 10^{-3})} \\\\\delta x \geq 1.85 \ \times \ 10^{-32} \ m[/tex]

Thus, the minimum uncertainty in its position is 1.85 x 10⁻³² m.