A bullet of mass 0.03 kg is fired from a rifle of mass 2kg,
which recoils with a velocity of 25.5m/s.
What is the velocity of the bullet?

The horizontal velocity (v) of the arrow when it left the bow is 39.1 m/s.

The time of motion of an object is the time taken for the object to travel from its initial position to the final position.

The time taken for the arrow to travel down the given height is calculated as follows;

h = vt + ¹/₂gt²

where;

h = ¹/₂gt²

2h = gt²

t² = 2h/g

t = √(2h/g)

t = √(2 x 9.5 / 9.8)

t = 1.39 seconds

The velocity (v) of the arrow when it left the bow is calculated as follows;

X = Vₓt

where;

Vₓ = X/t

Vₓ = (54.3 m) / (1.39 s)

Vₓ = 39.1 m/s

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The characteristics of electromagnetic radiation include amplitude (brightness), wavelength, frequency, and period. We have shown that the frequency of a light wave is proportional to its energy by the equation E = h v or E=hnu

so

Radiation with a wavelength of 8.1 X 10^-8 m has the highest energy

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The correct response that shows the forces that act on the ball is "all of the above."

Let us recall that a force is that which is able to impart acceleration to a body and also to change the direction of the object. We also have to know that from the Newton's law, force must always be able to produce an acceleration in an object.

As the child is sitting on the floor and is twirling the ball round the circle there are three forces that acts on the ball and all these force are given as;

1) Tension force

2) Gravitational force

3) Normal force

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No, law of conservation of energy is not applicable to open systems.

Law of conservation of energy states that energy present anywhere can neither be created nor be destroyed by any means, it can only be converted from one form of energy to other form of energy. It means total energy of an isolated system remains same.

An open system is a system that can exchange energy and matter from the surroundings. The flow of energy takes places in an open system. Therefore, the law of conservation does not apply to an open system as it can lose or gain energy, hence its total energy will not be constant and will violate law of conservation of energy.

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The magnitude of the gravitational force the moon exerts on Earth if Earth exerts a gravitational force of 1.99 x 10²⁰ newtons on the moon is 1.99 x 10²⁰ N

F = G m1 m2 / r²

F = Gravitational force of attraction

m1 = Mass of Earth

m2 = Mass of moon

r = Distance between moon and Earth

In this equation all the mentioned variables are constant for both gravitational force exerted by moon on Earth and by Earth on moon. So the value will be equal for both.

The equation mentioned above is the result of Newton's law of gravitation. This law states that any object in the universe attracts another object with a force which is directly proportional to product of both the masses and inversely proportional to the squared value of distance between them.

Therefore, the magnitude of the gravitational force the moon exerts on Earth is 1.99 x 10²⁰ N.

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The spring constant approximately is 1153 N/m.

When an 8 kg fish is suspended in the spring, its new length is 16.8 cm.

The details provided are

The spring's length is 10.0 cm (L1).

Fish mass, m = 4 kg

L2, the spring's final length, is 13.4 cm

The spring's extension is determined by the formula:

x = L2 - L1 x = 13.4 cm - 10.0 cm x = 3.4 cm

The following formula is used to get the spring's spring constant;

F = kx

mg = kx

k = mg/x = 4 * 9.8/0.034 = 1,152.94M/m

The new spring extension when an 8 kg fish is hung on it;

x = mg/k = 8 * 9.8/1152.94 = 0.068m = 6.8cm

6.8 cm + 10 cm = 16.8 cm is the spring's new length.

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The final volume of the balloon is 197,083 L.

They are proportional to each other if the pressure does not change. However, there are pitfalls. To perform calculations we have to use the so-called absolute temperature. Increased external air pressure compresses the balloon.

Using equation;

PV =nRT

here constants are P,n, R

so the relation between P and T is

V₁ /T₁= V₂/T₂

V₂ = (V₁ /T₁)×T₂

    = (125,000/300)×473

     = 197,083 L.

As the volume of the balloon decreases the pressure inside the balloon increases. After all external air pressure can compensate. It becomes a balloon filled with dense cold air. According to Boyle's law, the volume of a gas of constant mass at constant temperature is inversely proportional to its pressure. The air around you has weight and pushes away everything it touches. This pressure is called atmospheric pressure or barometric pressure. This is the force exerted on a surface by the air above as gravity pulls the surface toward the Earth.

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Based on the principle of moments, the correct options are:

1. If the outside support is removed, the plank will be balanced; option D

2. To balance the simple plank balance, place the 200 gram weight 125 cm left of the fulcrum; option C

3.  the mass of the mystery box is 8 kg; option D

The principle of moments is the principle that describes the turning effect produced by forces.

The principle of moments states that for a system of balanced forces, the sum of the clockwise moment about a point equals the sum of the anticlockwise moment about the same point.

The formula for calculating moments about a poit is given below:

Considering the given questions:

1. Assuming the fulcrum is the balance point of the plank balance.

Taking left as anticlockwise and right as clockwise motion

Anticlockwise moment = 30 * 3

Anticlockwise moment = 90

Clockwise moment = 45 * 2

Anticlockwise moment = 90

Therefore, the plank balance is balanced.

2. Assuming the fulcrum is the balance point of the plank balance.

Taking left as anticlockwise and right as clockwise motion

Clockwise moment = 500 * 50

Anticlockwise moment = 25000

Anticlockwise moment = 200 * d

Anticlockwise moment = 200d

200d = 25000

d = 125 cm to the left of the fulcrum

3. Assuming the fulcrum is the balance point of the plank balance.

Taking left as anticlockwise and right as clockwise motion

Clockwise moment = 10 * 4

Anticlockwise moment = 40

let the mass be m

Anticlockwise moment = m * 5

Anticlockwise moment = 5m

5m = 40

m = 8 kg

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The equivalent power of coal transported in a mile long train containing 120 cars, each holding 110 tons of coal travelling at 60 mph is 6.04 * 10e12 J

P = E / t

P = Power

E = Energy

t = Time

Distance of train = 1 mile

Number of cars in the train = 120

Tons of coal in 1 car = 110 tons

Velocity of train = 60 mph

Energy of one ton coal = 2.8×10e10 J

Total coal = 120 * 110

Total coal = 13200 tons

Total energy of coal = 13200 * 2.8×10e10

Total energy of coal = 369.6 * 10e12 J

v = d / t

v = Velocity

d = Distance

t = Time

t = d / v

t = 1 / 60

t = 0.017 h

t = 61.2 s

P = ( 369.6 * 10e12 ) / 61.2

P = 6.04 * 10e12 J

P = 6.04 pJ

Therefore, the equivalent power of coal is 6.04 * 10e12 J

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The information given is insufficient to compare the specific heat of the objects.

The specific heat capacity, commonly known as the specific heat capacity or cp in thermodynamics, is the product of the heat capacity of a sample of a substance and its mass. Informally, it is the quantity of heat that has to be added to a substance's mass in order to raise its temperature by a given amount. J kg 1 K, or specific heat capacity, is the SI unit for temperature. The specific heat capacity of water, for instance, is 4184 J kg 1 K1, or the amount of energy needed to elevate 1 kilogram of water by 1 K in temperature.

For each state of matter, specific heat capacity differs and frequently fluctuates with temperature.

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Answer:

D= 300 miles

t= 6 hours

Speed = [tex]\frac{300}{6}[/tex] = 50 miles/h

The speed of the car is 50miles/h.

[tex]\frac{50}{3}[/tex] * 5 km/h

= 83.34 km/h

The speed of the car is 83.34 km/h.

Explanation: The answer has been given in both miles per hour and kilometer per hour.

By using internal energy, the heat transferred to the air until its volume doubles is 5570.69 joule.

We need to know about internal energy to solve this problem. For a closed system, with matter transfer excluded, the changes in internal energy are due to heat transfer and due to thermodynamic work done by the system on its surroundings. It can be determined as

ΔU = Q - W

where ΔU is the change in internal energy, Q is heat transferred and W is work done.

From the question above, we know that

m = 6 kg

P1 = 200 kPa

T1 = 298 K

P2 = 400 KPa

The internal energy should be

ΔU = Q - W

m(U2 - U1) = Q2 - W2

at the second state the work is zero

m(U2 - U1) = Q2

Assume constant specific heats

Q2 = m . Cv (T2 - T1)

Using the ideal gas

P1/T1 = P2/T2

200/298 = 400/T2

T2 = 596 K

Q2 = m . C . (T2-T1)

Q2 = 6 . 3.1156 . (596-298)

Q2 = 5570.69 joule

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The velocity with which the collar strikes end B is 2.32 m/s.

How to calculate the work done by external forces on a body?
We conclude from the work energy theorem that any work performed on a body results in a change in total energy.

Given, mass of the collar = m = 0.5 kg
Therefore, effective weight of the collar = m*g = (0.5)*(9.81) = 4.905 N
Force acting on the collar = 5 N
We know, work done by external forces equates to the change in kinetic energy of the body.
Thus, ΔU = T ⇒ 0.5 * 0.5 * (v²-0) = 0.2[5 cos(30°) - 5 sin(30°) + 4.905)
⇒ v² = 5.39 m²/s² ⇒ v = 2.32 m/s.
Therefore, the velocity with which the collar strikes end B when acted upon by the 5N force is 2.32 m/s.

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The magnitude of the acceleration a of the astronaut during the throw is 0.2 m/s².

What is astronaut?

An astronaut is a person who has been prepared, outfitted, and sent into space as part of a human spaceflight program to serve as a commander or crew member. The phrase is occasionally used to refer to anyone who travels into space, including scientists, politicians, journalists, and tourists, despite the fact that it is typically designated for professional space travelers.

According to Newton's second law, an object's acceleration is inversely proportional to its mass and proportional to both the resulting force and the resultant force.

∑F = m. a

where, F = force, m = mass, a = acceleration of astronaut.
As per the information given in the question,

F = 16 N

M = 80 kg

putting these value in the equation,

16 = 80a

a = 16/80

a = 0.2 m/s²

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Answer:

D,Newton

Explanation:

b/c it is derived unit

Answer:

The newton

Explanation:

is drived because of its definition consist of multiplication of three defined base unit and nothing else it's not independently defined

When the coordinate increases at a rate of units per second, the rate of change of the distance from the particle to the origin at this instant would be 0.3 units per second.

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Answer:

The first two answers are true.

Since F = M a and there is no acceleration there is no net force acting on the puck.

With the use of formulas, it will take time 4.25 seconds for the object to reach the speed 41.65 m/s when dropped.

Speed is the distance travelled per time taken. It is a scalar quantity and it is measured in m/s

If an object is dropped and reached 41.65 m/s in its fall, Let us first know the height at which it was dropped by using the formula

v² = u² + 2as

Since the object was dropped, u = 0

41.65² = 0 + 2 × 9.8 × H

1734.72 = 19.6H

H = 1734.72/19.6

H = 88.51 m

To know how long it takes to reach this speed, let us use this formula

H = ut + 1/2gt²

where u = 0

88.51 = 1/2 × 9.8 × t²

88.51 = 4.9t²

t² = 88.51/4.9

t² = 18.06

t = √18.06

t = 4.25 s

Therefore, it takes 4.25 seconds of time to reach this speed.

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The tension force of the system is 395 N.

The acceleration of the system is 3.95 m/s².

The net force on the system is obtained by applying Newton's second law of motion as follows;

Net vertical force on the masses;

Fu = m₁g cosθ  + m₂g

where;

Fu = (100 kg x 9.8 m/s²) cos(25)  +  (100 kg x  9.8 m/s²)

Fu = 1,868.2 N

Apply the formula for Newton's second law of motion,

m₂a - Ff = -m₁a

where;

m₂a + m₁a = Ff

a(m₂ + m₁) = Fu sinθ

a = Fu sinθ / (m₂ + m₁)

a = (1868.2 x sin25) / (100 + 100)

a = 3.95 m/s²

The tension force is calculated as follows;

F(tens) = m₂a

F(tens) =  100 kg x 3.95 m/s² = 395 N

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The kinetic energy of the toy car after it has experienced the force is: 16 Joules

The kinetic energy formula and the procedure we will use is:

K.E. = 1/2 * m * v²

Where:

Information of the problem:

Applying kinetic energy formula we get:

K.E. = 1/2 * m * v²

K.E. = 1/2 * 2 kg (4 m/s)²

K.E. = 1/2 * 2 kg * 16 m²/s²

K.E. = 1/2 * 32 kg *m²/s²

K.E. =  16 kg*m²/s²

K.E. =  16 Joules

It is the energy possessed by a body due to its relative motion. It is usually expressed in Joules (J).

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The reaction force of the box will be 50 N in opposite direction.

Newton's third law of motion states that for every action there is equal and opposite reaction.

According to this law, the force exerted on an object is equal to the reaction experienced by the object.

The applied force is equal in magnitude but opposite in direction to the reaction of the impact.

Mathematically, this law can be illustrated as follows;

Fa = -Fbb

where;

Thus, we can conclude that if a box is pushed to the left by a certain force, the reaction force will be equal in magnitude to the applied force but opposite in direction.

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The torque experienced by a magnetic dipole moment M placed with its axis at angle theta with a uniform magnetic field B is τ = MB sin θ

Torque = Force * Perpendicular distance

F = m * B

F = Force

m = Magnetic pole strength

B = Magnetic field

τ = Torque

τ = m * B * 2 * l * sin θ

Magnetic dipole moment, M = 2 * m * l

τ = ( 2 * m * l ) * B * sin θ

τ = M B sin θ

Therefore, the torque experienced by a magnetic dipole moment M placed with its axis at angle theta with a uniform magnetic field B is τ = MB sin θ

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Answer:

[tex]0\; {\rm J}[/tex].

Explanation:

The speed of an object is a scalar quantity and is independent of the direction in which the object is moving.

The kinetic energy of an object depends only on the mass and speed of this object. If an object of mass [tex]m[/tex] is moving at a speed of [tex]v[/tex], the kinetic energy of this object will be [tex](1/2)\, m\, v^{2}[/tex].

As a result, as long as the speed of the object stays unchanged, changes in the direction of motion of the object will not affect the kinetic energy of the object.

In this question, the speed of this ball was initially [tex]1.72\; {\rm m\cdot s^{-1}}[/tex]. The kinetic energy of this object was initially [tex](1/2)\, (0.140\; {\rm kg})\, (1.72\; {\rm m\cdot s^{-1}})^{2} \approx 0.207\; {\rm J}[/tex].

The speed of this ball stays the same after the ball bounces back from the wall: [tex]1.72\; {\rm m\cdot s^{-1}}\![/tex]. As a result, the kinetic energy of this object would still be [tex](1/2)\, (0.140\; {\rm kg})\, (1.72\; {\rm m\cdot s^{-1}})^{2} \approx 0.207\; {\rm J}[/tex].

As a result, the change in the kinetic energy of this object would be [tex]0.207\; {\rm J} - 0.207\; {\rm J} = 0\; {\rm J}[/tex].

A 30MSUN main-sequence star has the greatest surface temperature.

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The particles of liquid are closely packed together than the particles of gas.

A liquid is a state of matter that contains particles that are loosely packed and that takes the shape of its container.

A gas is a state of matter that contains particles that are randomly placed and has no definite shape.

Temperature is the property of matter that defines how hot or cold a particular state of matter is.

When liquid and gas is being compared, the particles of the liquid is loosely packed together while the particles of gas are randomly placed and which collided freely with the walls of its container.

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When the velocity of Earth is doubled, the orbit of Earth around the Sun becomes smaller ( A ).

v = √ G M / r

v = Orbital velocity

G = Gravitational constant

M = Mass of the celestial body

r = Radius of the celestial body

Rearranging for r,

√ r = √ ( G M ) / v

Squaring on both sides,

r = G M / v²

If the velocity of Earth is doubled,

v' = 2v

r' = G M / v'²

r' = G M / ( 2 v )²

r' = G M / 4 v²

r' = r / 4

So the radius of orbit of Earth becomes one-fourth its initial radius.

Therefore, when the velocity of Earth is doubled, the orbit of Earth around the Sun becomes smaller ( A ).

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With the application of third equation of motion, the car go as far as 451.3 meters before coming to a stop

Braking Force is the force applied for an object to decelerate or stop. It is also the product of mass and its deceleration.

Given that a car with a mass of 1260 kg is travelling at 38.0 m/s when the driver sees a traffic jam ahead. If the average braking force in modern vehicles is 2010 N,

The given parameters are;

Before we know how far will the car go before coming to a stop, we need to know its deceleration.

F = ma

2010 = 1260a

a = 2010/1260

a = 1.6 m/s²

Now, let us applied third equation of motion

v² = u² - 2as

v = 0 since the car is coming to rest.

0 = 38² - 2 × 1.6 × s

0 = 1444 - 3.2s

3.2s = 1444

s = 1444/3.2

s = 451.25 m

Therefore, the car go as far as 451.3 meters before coming to a stop.

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Answer:

Approximately [tex]4.50 \times 10^{11}\; {\rm s}[/tex].

Explanation:

The speed of light is [tex]c \approx 3.00\times 10^{8}\; {\rm m\cdot s^{-1}}[/tex].

Note that the standard unit of energy, joule, is a derived unit. In terms of the standard base units:

[tex]\begin{aligned}1\; {\rm J} &= (1\; {\rm N})\, (1\; {\rm m}) \\ &= (1\; {\rm kg \cdot m \cdot s^{-2}})\, (1\; {\rm m}) \\ &= 1\; {\rm kg \cdot m^{2} \cdot s^{-2}} \end{aligned}[/tex].

Apply unit conversion and ensure that the unit of mass is in the standard unit kilogram ([tex]{\rm kg}[/tex]):

[tex]\begin{aligned} m &= 1\; {\rm g} \times \frac{1\; {\rm kg}}{10^{3}\; {\rm g}} &= 10^{-3}\; {\rm kg}\end{aligned}[/tex].

Apply the equation [tex]E = m\, c^{2}[/tex] to find the energy equivalent to [tex]m = 10^{-3}\; {\rm kg}[/tex] of matter:

[tex]\begin{aligned}E &= m\, c^{2} \\ &= (10^{-3}\; {\rm kg})\, (3.00 \times 10^{8}\; {\rm m\cdot s^{-1}})^{2} \\ &= 9.00\times 10^{13}\; {\rm kg \cdot m^{2} \cdot s^{-2}} \\ &= 9.00\times 10^{13}\; {\rm J} \end{aligned}[/tex].

Divide energy [tex]E[/tex] by power [tex]P[/tex] to find the duration [tex]t[/tex] of the power consumption:

[tex]\begin{aligned} t &= \frac{E}{P} \\ &\approx \frac{9.00 \times 10^{13}\; {\rm J}}{200\; {\rm J \cdot s^{-1}}} \\ &= 4.50 \times 10^{11}\; {\rm s}\end{aligned}[/tex].

In other words, the energy equivalent to [tex]1\; {\rm g}[/tex] of matter could power this fridge for approximately [tex]4.50 \times 10^{11}\; {\rm s}[/tex].