Using the momentum formula, calculate the mass of an object traveling at 3.1 m/s and has a momentum of 2365 kg m/s Dont forget your units.

Answer:

C) The book's inertia carried it forward.

When the bus stops suddenly, the book tends to remain in motion due to its inertia. The book was at rest on the seat of the bus, and when the bus stopped suddenly, the book continued moving forward with the same speed and direction it had before the bus stopped. As a result, the book slid off the seat and onto the floor.

Answer:

B) work-energy

Explanation:

It is the work-energy theorem as there is no theorem called "mass-acceleration theorem" [unless you consider Newton's Second Law (maybe?) but not really].

Its postulates are :

Answer:

80%

Explanation:

efficiency = (useful work out / total energy in) x  100

                = \(\frac{80}{100}\)  x 100

                =  80%

Answer:

21.31 meters

Explanation:

Since we're working with gravitational potential energy (GPE):

GPE (Joules) = mass (kg) * gravity (m/s^2) * height (meters)

1. Figure out what we have:

GPE = mass * gravity * height

We're looking for height, and we have the other three, so we're set to move on.

2. Isolate the unknown variable (height):

(GPE) / (mass * gravity) = height

3. Plug in your numbers:

(3.78 * 10^7 J) / ((1.81 * 10^5 kg) * (9.8 m/s^2)) = 21.31 meters

Answer:

\(kinetic \: energy = \frac{1}{2} m {v}^{2} \\ m = mass \: of \: the \: object \\ v = velocity \: \)

Answer:

2.5m/s^2

Explanation:

Note that the boat is reducing its speed. It is having negative acceleration or deceleration.

V^2 = u^2 -2as ( minus sign is used because of speed is reduced)

Given that,

s = 12 m

v = 5.5 m/s

u = 9.5 m/s

a = (v^2 - u^2) ÷ (-2s)

a= ( 5.5^2 - 9.5^2) ÷ ( -2× 12)

a = 2.5 m/s^2

The acceleration of the boat is -2.5 m/s²

The given paramters;

distance traveled by the boat, d = 12 m

initial velocity of the boat, u = 9.5 m/s

final velocity of the boat, v = 5.5 m/s

The acceleration of the boat is calculated as;

\(v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(5.5)^2 -(9.5)^2 }{2(12)} \\\\a = -2.5 \ m/s^2\)

Thus, the deceleration of the boat is 2.5 m/s²

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The radius of curvature of the sphere is 0.397 cm.

Refractive index of the liquid, n₁ = 1.3

Refractive index of the glass, n₂ = 1.5

Distance of object away from the sphere, d₁ = 40 cm

Distance of object in front of the sphere, d₂ = 32 cm

According to the equations of refraction in glass sphere,

n₁/d₁ + n₂/d₂ = (n₂ - n₁)/R

Applying the values of n₁, n₂, d₁ and d₂,

(1.3/40) + (1.5/32) = (1.5 - 1.3)/R

0.0794 = 0.2R

Therefore, the radius of curvature of the sphere,

R = 0.0794/0.2

R = 0.397 cm

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The shorter the distance between the balls, the greater the electric force between them.

option C

The electric force between charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

F = kq²/r²

where;

Therefore, if the charges on the pith balls are fixed, the closer they are, the stronger the electric force between them will be. So option C is the correct answer.

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Answer:

B. 10 cm

x^3 = 1 L

1 L = 1000 cm^3 (since 1 L = 1000 cm^3)

x^3 = 1000 cm^3

Answer:

A systematic error may result in a high degree of precision.

A systematic error will likely result in poor accuracy

Explanation:

It's what my work gave me when I got it wrong

Explanation:

The density of earth \(\rho_E\) is given by

\(\rho_E = \dfrac{M_E}{\left(\frac{4\pi}{3}R_E^3\right)}\)

and in terms of this density, we can write the acceleration due to gravity on earth as

\(g_E =G\dfrac{M_E}{R_E^2} = \dfrac{4\pi G}{3}\rho_ER_E\)

Similarly, the acceleration due to gravity \(g_P\) on this new planet is given by

\(g_P = G\dfrac{M_P}{R_P^2} = G\dfrac{\frac{4\pi}{3}R_p^3\rho_P}{R_P^2}\)

\(\:\:\:\:\:= \dfrac{4\pi G}{3}\rho_PR_P\)

We know that this planet has the same density as earth and has a radius 2 times as large. We can then rewrite \(g_P\) as

\(g_P = \dfrac{4\pi G}{3}\rho_E(2R_E)\)

\(\:\:\:\:\:= 2\left(\dfrac{4\pi G}{3}\rho_ER_E\right) = 2g_E\)

\(\:\:\:\:\:= 2(9.8\:\text{m/s}^2) = 19.6\:\text{m/s}^2\)

Answer:

This link was diagram

Explanation:

https://doubtnut.app.link/FnsNC80Dccb

Based on Newton's first  and second law of motion most people would find it less painful to catch a flying baseball than a bowling ball flying at the same speed as the baseball because the mass of the baseball is smaller and will require smaller force to be stopped.

Newton's first law of motion first law of motion states that a body at rest or uniform motion in a straight line will continue in that path unless acted upon by an external force.

Newton's first law of motion is also called law of inertia because it depends on mass of the object.

An object with a greater mass will require greater force to be stopped or get moving.

Based on Newton's first law of motion most people would find it less painful to catch a flying baseball than a bowling ball flying at the same speed as the baseball because the mass of the baseball is smaller and will require smaller force to be stopped.

Also according to Newton's second law of motion, the force applied to an object is proportional to the product of mass and acceleration of the object. Thus, a baseball with smaller mass will require smaller force to be stopped.

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Given that,

Charge = 16 nC

Mass = 33 g

Distance = 27 cm

We need to calculate the acceleration

Using formula of electrostatic force

\(F=\dfrac{kqQ}{r^2}\)

\(ma=\dfrac{kq^2}{r^2}\)

Put the value into the formula

\(33\times10^{-3}\times a=\dfrac{9\times10^{9}\times(16\times10^{-9})^2}{(27\times10^{-2})^2}\)

\(a=\dfrac{9\times10^{9}\times(16\times10^{-9})^2}{(27\times10^{-2})^2\times33\times10^{-3}}\)

\(a=0.0009577\ m/s^2\)

\(a=9.577\times10^{-4}\ m/s^2\)

We need to calculate the speed of charge

Using equation of motion

\(v^2=u^2+2as\)

Where, v= speed

u = initial speed

a = acceleration

s = distance

\(v^2=0+2\times9.577\times10^{-4}\times16\times10^{-2}\)

\(v=\sqrt{0+2\times9.577\times10^{-4}\times16\times10^{-2}}\)

\(v=0.0175\ m/s\)

Hence, The speed of the charge is 0.0175 m

Answer:

trial and error.

Explanation:

a problem solving method that involves trying all possible solutions until one works is using trail and error.

(a) The work done by the donkey on the cart is 59,721.9 J.

(b) The work done by the force of gravity on the cart is -48,434.87 J.

(c) The work done on the cart by friction during this time is 11,315.12 J.

(a) The work done by the donkey on the cart is calculated as follows;

Wd = Fd cosθ

where;

Wd = 375 N x 163 m x cos(12.3)

Wd = 59,721.9 J

(b) The work done by the force of gravity on the cart is calculated as;

Wg =  Fg x d x cosθ

Where;

θ = 90⁰ + 4.03⁰ = 94.03⁰

Wg = (431 kg x 9.81 m/s²) x 163 m x cos (94.03)

Wg = -48,434.87 J

(c) The work done on the cart by friction during this time is calculated as;

Wf = Ff x d x cosθ

where;

Ff = μmg cosθ

Ff = 0.0165 x 431 kg x 9.81 x cos (4.03)

Ff = 69.59 N

Wf = 69.59 x 163 x cos (4.03)

Wf = 11,315.12 J

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Answer:

when air or water cools it sinks

The volume of the gas at 101.3 kPa would be approximately 651.25 cm³.

To calculate the change in volume of a gas from an initial pressure to a final pressure, we can use Boyle's law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.

Boyle's law can be expressed as:

P1 * V1 = P2 * V2

Where:

P1 = Initial pressure (80.8 kPa)

V1 = Initial volume (817 cm³)

P2 = Final pressure (101.3 kPa)

V2 = Final volume (to be calculated)

Let's plug in the values into the equation and solve for V2:

80.8 kPa * 817 cm³ = 101.3 kPa * V2

V2 = (80.8 kPa * 817 cm³) / 101.3 kPa

V2 ≈ 651.25 cm³

Therefore, the volume of the gas at 101.3 kPa would be approximately 651.25 cm³.

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Double

Explanation:

Since the period T of a pendulum is given by

\(T = 2\pi \sqrt{\dfrac{l}{g}}\)

By increasing the length of the pendulum by 4, the period becomes

\(T' = 2\pi \sqrt{\dfrac{4l}{g}} = 2\left(2\pi \sqrt{\dfrac{l}{g}}\right) = 2T\)

You can see that the period doubles when we increase the length by a factor of 4.

Block A is moving at around speed is 7.91 m/s right before it hits the ground.

The ball's parabolic motion causes it to move at a speed of 26.3 m/s right before it strikes the ground, which is faster than its straight-downfall velocity of 17.1 m/s.

Now, let's calculate the system's potential energy when collar C is removed:

PE = mgh

where

m = 4 kg (mass of block A)

g = 9.81 m/s² (acceleration due to gravity) (acceleration due to gravity)

h = 0.6 m (height that block A is raised)

PE = 4 kg × 9.81 m/s² × 0.6 m

PE = 23.54 J

Let's now find the kinetic energy of the system just before block A hits the ground:

KE = (1/2) × m × v²

where

m = 4 kg (mass of block A)

v = the block A's speed right before impact with the ground.

KE = (1/2) × m_A × v² + (1/2) × m_B × v²

where

m_A = 4 kg (mass of block A)

m_B = 5 kg (mass of block B)

d_total = d_A + d_B

where

d_A = 0.6 m (distance that block A moves)

d_B = 1 m (distance that block B moves)

d_total = 0.6 m + 1 m

d_total = 1.6 m

PE = KE

mgh = (1/2) × m_A × v² + (1/2) × m_B × v²

Substituting the known values, we get:

4 kg × 9.81 m/s² × 0.6 m = (1/2) × 4 kg × v² + (1/2) × 5 kg × v²

14112 J = (4.5 kg) × (1/2) × v²

v² = 62.693 m²/s²

v = 7.91 m/s.

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The law of conservation of energy is a general principle that applies to all closed systems, while the first law of thermodynamics is a specific application of this principle to thermodynamic systems.

The law of conservation of energy and the first law of thermodynamics are related, but they are not exactly the same thing.

The law of conservation of energy states that the total energy of a closed system remains constant over time. Energy can be transformed from one form to another (such as from potential to kinetic energy), but the total amount of energy in the system remains constant. This is a fundamental law of physics and applies to all systems, not just thermodynamic ones.

The first law of thermodynamics is a specific application of the law of conservation of energy to thermodynamic systems. It states that the total energy of a thermodynamic system (including its internal energy, potential energy, and kinetic energy) is constant, but energy can be transferred into or out of the system in the form of heat or work. In other words, the first law of thermodynamics is a statement about the relationship between energy and work in thermodynamic processes.

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Answer:

As S= Vt

S = (23.27)(6.08)

S = 141.48 m

Explanation:

For work to be considered done by a physicist, it must involve the application of physics principles and concepts to solve a problem or answer a research question. This can involve theoretical work, such as mathematical modeling, or experimental work, such as designing and conducting experiments to test hypotheses.

An example of work done by a physicist would be the design and analysis of experiments to study the properties of a new material. This could involve developing a theoretical model to predict the behavior of the material, designing and building experimental apparatus, conducting experiments, and analyzing the resulting data using statistical and mathematical techniques. The physicist would use their knowledge of physics principles and concepts to interpret the data and draw conclusions about the properties of the material.

Answer:

Approximately \(71.6\; {\rm N}\) (assuming that \(g = 9.81\; {\rm N \cdot kg^{-1}}\).)

Explanation:

Refer to the diagram attached. Forces on this object are:

Assuming that \(g = 9.81\; {\rm N \cdot kg^{-1}}\), the magnitude of the weight of the sign would be:

\(\begin{aligned}(\text{weight}) &= m\, g \\ &= (7.30\; {\rm kg})\, (9.81\; {\rm N\cdot kg^{-1}}) \\ &\approx 71.6\; {\rm N}\end{aligned}\).

Note that weight points downwards (negative) and is entirely in the vertical direction. As a result, the \(y\)-component of weight would be equal to \((-71.6)\; {\rm N}\).

Hence, the \(y\)-component of these forces would be:

Since forces on the object to be balanced, forces need to be balanced in each component. For forces in the \(y\)-component to be balanced, forces in the vertical direction need to add up to \(0\; {\rm N}\):

\(0\; {\rm N} + (-71.6)\; {\rm N} + (\text{$y$-component of tension on the right}) = 0\; {\rm N}\).

Hence, the \(y\)-component of the tension from the wire on the right end would be \(71.6\; {\rm N}\).

Answer:

Period is the time taken for a complete cycle or oscillation or revolution.

Symbol: T

Units: seconds

Frequency is the number of complete oscillations.

Symbol: f

Units: Hertz ( Hz )

Relationship

Explanation:

\(T = \frac{1}{f} \)

Answer:

Its speed remains the same but its velocity keeps chaging (in direction)

Explanation:

GREETINGS!

Answer is 553 m/s

By the formula of,

       Vf = Vi + at

  where,

             Vi is initial velocity

             a is acceleration due to gravity

             t is time.

As we know that,

                            F = ma

so,

                   \(a = \frac{F}{m}\)

put this in the final velocity formula,

                  Vf = Vi + (F/m)t

                  Vf = 125 + (25000/440)7.5

                  Vf = 551.13 m/s

HOPE IT HELPS!

Answer:

Explanation:

a) the acceleration of the puck on the rough ice.

a = μg = 0.550(9.81) = 5.3955 = 5.40 m/s²

 (comes from μ = F/N = ma/mg = a/g)

b) the distance from the end boards the puck is when it comes to a stop.

v² = u² + 2as

0² = 12.0² + 2(-5.40)s

s = 13.3 ft

so distance from the boards is

15.7 - 13.3 = 2.4 m

by the way...that's some VERY rough ice...more like sand.

(D)

Explanation:

The more massive an object is, the greater is the curvature that they produce on the space-time around it.

The theory of relativity explain the gravity exerted by massive objects is

more massive objects create larger curves of space-time (option-d).

The term "gravitational force" refers to the attraction between masses. The gravitational force increases in size as the masses get bigger (also called the gravity force). As the distance between masses grows, the gravitational force progressively lessens.

Greater gravitational forces will be used to attract heavier things since the gravitational force is directly proportional to the mass of both interacting objects. Therefore, when two things' respective masses increase, so does their gravitational pull to one another.

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