Help pleaseeee need the answers ASAP

Answer:

3.44× 10⁻¹⁹Joules

Explanation:

Energy of the wavelength is expressed using the formula:

E = hc/λ

h is the Planck constant

c is the velocity of light

λ is the wavelength

Given

h = 6.63 × 10^-34 m² kg / s

c = 3×10⁸ m/s

λ = 579nm = 579 × 10⁻⁹m

λ = 5.79× 10⁻⁷m

Substitute the given values into the formula

E = hc/λ

E = (6.63 × 10⁻³⁴× 3×10⁸)/5.79× 10⁻⁷

E = 19.89× 10⁻³⁴⁺⁸/5.79× 10⁻⁷

E = 19.89× 10⁻²⁶/5.79× 10⁻⁷

E = 3.44× 10⁻²⁶⁺⁷

E = 3.44× 10⁻¹⁹Joules

Hence the energy of this wavelength of light is 3.44× 10⁻¹⁹Joules

Answer:

2.3 newtons of force

Explanation:

Divide the weight by the speed of the bike

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:

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:

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.

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Answer:

c. joules per gram degree Celsius

Explanation:

edg 2021

Answer:

a refractor telescope

Explanation:

Answer:

a. refractor telescope

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:

[tex]\huge\boxed{\sf P.E. = 39.2\ Joules}[/tex]

Explanation:

Given Data:

Mass = m = 4 kg

Acceleration due to gravity = g = 9.8 m/s²

Height = h = 1 m

Required:

Potential Energy = P.E. = ?

Formula:

P.E. = mgh

Solution:

P.E. = (4)(9.8)(1)

P.E. = 39.2 Joules

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

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

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:

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.

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Answer:

The answer is Spring Potential

Explanation:

Because all the others are a mechanical form of energy

Answer:

Distance and time.

Explanation:

Speed=Distance/time

The two factors which we need in order to calculate the speed of an object are the distance covered by the object and the time taken to cover that distance.

Speed is the rate of change of position of an object in any direction. Speed is a scalar quantity as it has only magnitude and no direction. It is measured as the ratio of the distance covered by an object to the time taken in which the distance was covered by that object.

Speed has the dimension of distance covered by the time taken. Thus, the SI unit of speed is the combination of the basic units of distance and the basic unit of Time. Thus, the SI unit of speed is meter per second (m/s).

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Answer:

4770 N

Explanation:

Momentum is the product of mass and velocity and force is the change in momentum divided by change in time.

Given from the question;

Mass of arrow= 0.53 kg

Velocity of arrow = 63 m/s

Initial velocity of arrow = 0 m/s

Change in time = 0.007 s

Finding momentum after the arrow is released as;

p=m*v

p= 0.53 * 63

p= 33.39 kg*m/s

Force is the change in momentum divided by change in time;

F= 33.39 / 0.007

F= 4770 N

Answer:

4.44 m/s^2

Explanation:

A 15-kg wagon is pulled to the right across a surface by a tension of 100 newtons at an angle of 30 degrees above the horizontal. A frictional force of 20 newtons to the left acts simultaneously.The acceleration of the wagon 4.44 m/s².

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.

Given in the question a 15 kg wagon is pulled to the right across a surface by a tension of 100 newtons at an angle of 30 degrees above the horizontal. A frictional force of 20 newtons to the left acts simultaneously, then the acceleration is  4.44 m/s².

The acceleration of the wagon 4.44 m/s².

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Answer:

zero

Explanation:

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

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

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:

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:

If the heavy load had been lifted more slowly, the work done on the load would have been the same.

Explanation:

Work done on an object is given as;

W = Fd

where;

F is the force applied on the object

d is the displacement of the object

for the given question, the applied force on the load = mg (mass of the load multiplied by acceleration due to gravity).

Also, the displacement of the object = vertical height the load was lifted.

W = mgh

The work done on the load is independent of time.

Thus, if the heavy load had been lifted more slowly, the work done on the load would have been the same.

A person lifting a heavy load to a vertical height of 2.0 m in 3 seconds does the same work as if he/she lifts it in 6 s.

A person lifts a heavy load to a vertical height of 2.0 m in 3 seconds.

We want to compare the work done with the one that he/she would have done if the process had taken 6 seconds.

In physics, work (W) is the energy transferred to or from an object via the application of force (F) along a displacement (s).

W = F × s

Given the displacement is the same (2.0 m) and the force needed is also the same (weight of the object), the work is the same for both processes.

A person lifting a heavy load to a vertical height of 2.0 m in 3 seconds does the same work as if he/she lifts it in 6 s.

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Answer:

a) The diagrams can be seen in the picture attached

(b) By comparing the diagrams we can conclude that the resultant R₁ = R₂ = R₃

Further explanation

Vector is quantity that has magnitude and direction.

One example of a vector is acceleration.

Acceleration is rate of change of velocity.

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem !

This problem is about Vector and Vector Diagram.

Given:

Vector A = -200 j

Vector B = -250 i

Vector C = (150 sin 30.0°) i + (150 cos 30.0°) j = 75 i + 75√3 j

Unknown:

R₁ = A + B + C = ?

R₂ = B + C + A = ?

R₃ = C + B + A = ?

Solution:

R₁ = A + B + C  = (-200 j) + (-250 i) + (75 i + 75√3 j)

R₁ = -175i + (75√3 - 200)j

R₂ = B + C + A = (-250 i) + (75 i + 75√3 j) + (-200 j)

R₂ = -175i + (75√3 - 200)j

R₃ = C + B + A  = (75 i + 75√3 j) + (-250 i) + (-200 j)

R₃ = -175i + (75√3 - 200)j

From the results above, it can be concluded that the resultants above produce the same results. This can be confirmed from the diagrams in the attachment.

Explanation:

Answer:

y = 14238 m.    the height of the rocket is much less than this distance therefore the plan will not work.

Explanation:

Let's analyze this exercise, so that the Martian's plan works, the vertical height of the body must be zero when it is more than half of the way to the planet Mars, this is so that Mars attracts it and can arrive.

Let's calculate the maximum height of the launch

          [tex]v_{y} ^2 = v_{oy}^2 - 2 g y[/tex]

at the highest point [tex]v_{y}[/tex] = 0

          y = v_{oy}² / 2g

          y = (v₀ sin θ)² / 2g

let's calculate

          y = (1250 sin 25)² /2 9.8

          y = 14238 m

In the exercise, indicate that the distance to Mars is h = 321770 m, half of this distance is

          h / 2 = 160885 m

therefore the height of the rocket is much less than this distance therefore the plan will not work.

The height reached is low, so it is not necessary to take into account the variation of g with height