What is the solution?

Answer:

The one with the faster velocity is the one with a velocity of -10m/s

Answer:

During a phase change, a substance undergoes transition to a higher energy state when heat is added, or to a lower energy state when heat is removed. Heat is added to a substance during melting and vaporization. Latent heat is released by a substance during condensation and freezing. Explanation:

Answer:

B. 450 N

Explanation:

Use Hooke's law:

F = kx

150 N = k (1 m)

k = 150 N/m

F = kx

F = (150 N/m) (3 m)

F = 450 N

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 40 × 4

We have the final answer as

Hope this helps you

Answer:68.15m/s

Explanation:

Given:

v₁=15m/s

a=6.5m/s²

v₁=?

x=340m

Formula:

v₁²=v₁²+2a (x)

Set up:

=[tex]\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)[/tex]

An open cluster is a group of up to a few thousand stars that were formed from the same giant molecular cloud, and are still loosely gravitationally bound to each other. In contrast, globular clusters are very tightly bound by gravity. ... Open clusters are very important objects in the study of stellar evolution.

Explanation:

When sound travels in a given "medium", it would touch the surface of another "medium" and will bounce back in some other direction, this occurrence is called the reflection of sound.

Answer:

2nd option on edge2021

Explanation:

Answer: Calculate the energy required in joules to raise the temperature of 450 grams of water from 15°C to 85°C? (The specific heat capacity of water is 4.18 J/g/°C)

Explanation:

The quantity of heat is required to raise the temperature of of water is 94050 joule.

Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.

The law of energy conservation is adhered to by all energy forms. The law of conservation of energy essentially says that the total energy of the system is conserved in a closed system, also known as an isolated system.

Mass of water: m = 450 grams

Initial temperature of water = 35° C

Final temperature of water = 85° C

Capacity of water is 4.18 J/g °C.

Hence, the  quantity of heat is required to raise = mass × Capacity  × raise in temperature

= 450 × 4.18 × (85 - 35) joule

= 94050 joule.

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Answer:

2156 N

Explanation:

Data obtained from the question include:

Mass of satellite (m) = 220 Kg

Force (F) of gravity =?

The force of gravity exerted on the satellite on the surface of the earth can be obtained by using the following formula:

Force (F) of gravity = mass (m) × acceleration due to gravity (g)

F = mg

Mass of satellite (m) = 220 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) of gravity =?

F = mg

F = 220 × 9.8

F = 2156 N

Thus, the force of gravity exerted on the satellite on the surface of the earth is 2156 N

Answer:

The two charged objects will exert equal and opposite forces on each other.

Explanation:

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of charges on the objects and inversely proportional to the square of the distance between the two objects.

This force of attraction or repulsion between the two charged objects is always equal and opposite.

Therefore, the two charged objects will exert equal and opposite forces on each other.

Answer:

[tex]E_s = 277.13V[/tex]

Explanation:

Given

[tex]Load\ Voltage = 480V[/tex]

Required

Determine the voltage dropped in each stage.

The relation between the load voltage and the voltage dropped in each stage is

[tex]E_l = E_s * \sqrt3[/tex]

Where

[tex]E_l = 480[/tex]

So, we have:

[tex]480 = E_s * \sqrt3[/tex]

Solve for [tex]E_s[/tex]

[tex]E_s = \frac{480}{\sqrt3}[/tex]

[tex]E_s = \frac{480}{1.73205080757}[/tex]

[tex]E_s = 277.128129211[/tex]

[tex]E_s = 277.13V[/tex]

Hence;

The voltage dropped at each phase is approximately 277.13V

v² - u² = 2 a ∆x

where u = initial velocity, v = final velocity, a = acceleration, and ∆x = distance traveled.

So

v² - (15 m/s)² = 2 (6.5 m/s²) (340 m)

v² = 4645 m²/s²

v ≈ 68.15 m/s

Answer:

30 m/s

Explanation:

v = u + at

given that,

v = 10 + ( 2 × 10)

v = 10 + 20

v = 30 m/s

Answer:

I think it will get colder

Explanation:

Answer:

The water molecules go faster as it gets colder they go slower

Explanation:

trust me thats the answer

Answer:

Average speed

Explanation:

250/50=5

1000/20=5

Answer:

Written in Python

def energyvector(mass):

    c = 2.9979 * 10**8

    energy = mass * c ** 2

    print(round(energy,2))

Explanation:

This line defines the function

def energyvector(mass):

This line initializes the speed of light

    c = 2.9979 * 10**8

This line calculates the corresponding energy

    energy = mass * c ** 2

This line prints the calculated energy

    print(round(energy,2))

Answer:

913 hours ur welcome :)

Answer:

  φ = B sin (2π n/a   x)

Explanation:

In quantum mechanics when a particle moves freely it implies that the potential is zero (V = 0), so its wave function is

     φ = A cos kx + B sin kx

we must place the boundary conditions to determine the value of the constants A and B.

In our case we are told that the particle cannot be outside the boundary given by x = ± a / 2

therefore we must make the cosine part zero, for this the constant A = 0, the wave function remains

    φ = B sin kx

the wave vector is

      k = 2π /λ

now let's adjust the period, in the border fi = 0 therefore the sine function must be zero

         φ (a /2) = 0

          0 = A sin (2π/λ  a/2)

therefore the sine argument is

          2π /λ   a/2 = n π

          λ= a / n

we substitute

          φ = B sin (2π n/a   x)

Answer:

(a). The kinetic friction force is 1950 N.

(b). The magnitude of force will be equal of friction force

Explanation:

Given that,

Constant speed = 6.38 m/s

Force [tex]F=7.50\times10^{3}\ N[/tex]

Kinetic friction = 0.26

(a). We need to calculate the friction force

Using formula of friction force

[tex]f_{k}=\mu F_{N}[/tex]

Put the value into the formula

[tex]f_{k}=0.26\times7.50\times10^{3}[/tex]

[tex]f_{k}=1950\ N[/tex]

(b). If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh,

We need to calculate the magnitude of this force

According to given data,

The same force will be applied to keep constant velocity.

Hence, (a). The kinetic friction force is 1950 N.

(b). The magnitude of force will be equal of friction force.

(a). The kinetic friction force is 1950 N.

(b). The magnitude of force will be equal of friction force

a. The magnitude of the kinetic friction force experienced by the sleigh is

[tex]= 0.76 \times 7.50 \times 10^3[/tex]

= 1950 N

b. It should be equivalent to the friction force.

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Answer:

The  value of the power is   [tex]P_c  =  38.55 \  W [/tex]

Explanation:

From the question we are told that

   The  power  rating [tex]P_{1000} =P_b=  52 \  W[/tex]

    The frequency is  [tex]f = 1000 \  Hz[/tex]

    The  frequency at which the sound intensity decreases  [tex]f_k  =  20 \  Hz[/tex]

     The decrease in intensity is by [tex]\beta  =  1.3 dB[/tex]

Generally the  initial intensity of the speaker  is mathematically represented as

     [tex]\beta_1 =  10 log_{10} [\frac{P_b}{P_a} ][/tex]

Generally the intensity of the speaker after it has been decreased is

       [tex]\beta_2 =  10 log_{10} [\frac{P_c}{P_a} ][/tex]

So

[tex]\beta_1-\beta_2 =  10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ][/tex]

=>  [tex]\beta =  10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3[/tex]

=>  [tex]\beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3[/tex]

=>  [tex]\beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3[/tex]

=> [tex]10log_{10} [\frac{P_b}{P_c} ] = 1.3[/tex]

=> [tex]log_{10} [\frac{P_b}{P_c} ] = 0.13[/tex]

taking atilog of both sides

[tex][\frac{P_b}{P_c} ] = 10^{0.13}[/tex]      

=>[tex][\frac{52}{P_c} ] = 10^{0.13}[/tex]      

=>  [tex]P_c  =  \frac{52}{1.34896}[/tex]

=>   [tex]P_c  =  38.55 \  W [/tex]

Answer:

35.35 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =.?

The range (i.e how far away) of the ball can be obtained as follow:

R = u² Sine 2θ /g

R = 20² Sine (2×30) / 9.8

R = 400 Sine 60 / 9.8

R = (400 × 0866) / 9.8

R = 346.4 / 9.8

R = 35.35 m

Therefore, the range (i.e how far away) of the ball is 35.35 m

Answer:

(a) The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.

(b) The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.

(c) Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.

(d) The work done by the mattress on the bowling ball is 113.272 joules.

Explanation:

The statement is incomplete. The complete question is:

In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.

(a) What is the kinetic energy of the ball just before it hits the mattress?  

(b) How much work does the gravitational force of the earth do on the ball as it falls, for the first part of the fall (from the moment you drop it to just before it hits the mattress)?  

(c) How much work does the gravitational force do on the ball while it is compressing the mattress?

(d) How much work does the mattress do on the ball? (You’ll need to use the results of parts (a) and (c)）

(a) Based on the Principle of Energy Conservation, we know that ball-earth system is conservative, so that kinetic energy is increased at the expense of gravitational potential energy as ball falls:

[tex]K_{1}+U_{g,1} = K_{2}+U_{g,2}[/tex] (Eq. 1)

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Kinetic energies at top and bottom, measured in joules.

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Gravitational potential energies at top and bottom, measured in joules.

Now we expand the expression by definition of gravitational potential energy:

[tex]U_{g,1}-U_{g,2} = K_{2}-K_{1}[/tex]

[tex]K_{2}= m\cdot g \cdot (z_{1}-z_{2})+K_{1}[/tex] (Eq. 1b)

Where:

[tex]m[/tex] - Mass of the bowling ball, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the bowling ball, measured in meters.

If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex], [tex]z_{2} = 0\,m[/tex] and [tex]K_{1} = 0\,J[/tex], the kinetic energy of the ball just before it hits the matress:

[tex]K_{2} = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1.5\,m-0\,m)+0\,m[/tex]

[tex]K_{2} = 102.974\,J[/tex]

The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.

(b) The gravitational work done by the gravitational force of Earth ([tex]\Delta W[/tex]), measured in joules, is obtained by Work-Energy Theorem and definition of gravitational potential energy:

[tex]\Delta W = U_{g,1}-U_{g,2}[/tex]

[tex]\Delta W = m\cdot g\cdot (z_{1}-z_{2})[/tex] (Eq. 2)

If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex] and [tex]z_{2} = 0\,m[/tex], then the gravitational work done is:

[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.5\,m-0\,m)[/tex]

[tex]\Delta W = 102.974\,J[/tex]

The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.

(c) The work done by the gravitational force of Earth while the bowling when mattress is compressed is determined by Work-Energy Theorem and definition of gravitational potential energy:

[tex]\Delta W = U_{g,2}-U_{g,3}[/tex]

Where [tex]U_{g,3}[/tex] is the gravitational potential energy of the bowling ball when mattress in compressed, measured in joules.

[tex]\Delta W = m\cdot g \cdot (z_{2}-z_{3})[/tex]

Where [tex]z_{3}[/tex] is the height of the ball when mattress is compressed, measured in meters.

If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2}= 0\,m[/tex] and [tex]z_{3} = -0.15\,m[/tex], the work done is:

[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0\,m-(-0.15\,m)][/tex]

[tex]\Delta W = 10.298\,J[/tex]

Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.

(d) The work done by the mattress on the ball equals the sum of kinetic energy just before mattress compression and the work done by the gravitational force when mattress is compressed:

[tex]\Delta W' = K_{2}+\Delta W[/tex]

([tex]K_{2} = 102.974\,J[/tex], [tex]\Delta W = 10.298\,W[/tex])

[tex]\Delta W' = 113.272\,J[/tex]

The work done by the mattress on the bowling ball is 113.272 joules.