a projectile is fired into a region of no air resistance the horizontal component of velocity remains constant because projectiles
A) is not acted on by any horizontal force
B) is affected by gravity
C) has no vertical component of velocity

Thermal energy is transferred between objects only when they have different temperatures.

The movement of atoms and molecules within a substance is a form of energy known as thermal energy.

When two objects are in contact, or even in proximity, the atoms and molecules in the hotter object collide with those in the colder object.

Because of these effects, a portion of the kinetic energy from the more sizzling object is moved to the colder object, raising the temperature of the previous and bringing down the temperature of the last option.

A material's thermal conductivity — an estimation of how well it moves heat — decides how rapidly heat is moved between various sorts of materials.

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It takes approximately 0.902 seconds time interval for the ball to strike the ground.

To determine the time interval it takes for the ball to strike the ground, we can use the equations of motion. Since the ball is thrown directly downward, we can assume that the initial velocity (u) is negative (-8.85 m/s), and the acceleration due to gravity (g) is positive (9.8 m/s²).

The equation to calculate the time of flight (t) is

t = (v - u) / g

Where:

v = final velocity (when the ball strikes the ground)

u = initial velocity

g = acceleration due to gravity

In this case, the final velocity when the ball strikes the ground is 0 m/s (as it comes to rest). So, substituting the given values into the equation, we have

t = (0 - (-8.85)) / 9.8

t = 8.85 / 9.8

t = 0.902 seconds

Therefore, it takes approximately 0.902 seconds for the ball to strike the ground.

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Substituting the values you get that v is 2.18 x 106 ms-1.

(D)

A concave lens is thinner in the middle than it is at the edges. This causes parallel rays to diverge

now, in the options we have

A plano concave lens is an optical lens with one concave surface and one flat surface. It has a negative focal length,

so, the answer is

(D)

I hope this helps you

Answer:

F = 30 N

Explanation:

Given that,

Mass of a toy train, m = 1.5 kg

The acceleration of the toy-train, a = 20 m/s²

We need to find the force acting on it. We know that, net force acting on an object is given by :

F = ma

Substitute all the values,

F = 1.5 × 20

F = 30 N

Hence, the net force acting on the toy train is equal to 30 N.

Answer: 4g

Explanation:

Given

The time on earth is 4 s for 20 oscillations

i.e. time period for one oscillation is \(\frac{4}{20}=\frac{1}{5}\ s=0.2\ s\)

for planet X it is 40 oscillations in 4 s i.e.

time period is \(\frac{4}{40}=0.1\ s\)

The time period of a pendulum is given by

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

where L=length of the pendulum

g=acceleration due to gravity

So, we can write the ratio of the time period of two pendulums as

\(\frac{0.2}{0.1}=\dfrac{2\pi\sqrt{\frac{L}{g}}}{2\pi\sqrt{\frac{L}{g'}} }\)

\(2=\sqrt{\frac{g'}{g}}\)

\(\sqrt{g'}=2\sqrt{g}\\g'=4g\)

i.e. 4 times the gravity of the earth

According to the giant impact hypothesis about the formation of the moon, the earth did not break apart into many pieces when the giant impact happened because "the impactor was about the size of Mars or smaller, so it ejected material from the earth but did not break it."

The giant impact hypothesis proposes that the Moon was formed out of debris ejected by a Mars-sized object (named Theia) that struck the early Earth almost 4.5 billion years ago.

The collision created a disk of material that eventually coalesced to form the Moon. Therefore, the correct option is that the impactor was about the size of Mars or smaller, so it ejected material from the earth but did not break it.

This means that the impactor was too small to completely destroy the Earth, but it was enough to disintegrate it and create a large debris disk that eventually formed the Moon.

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Answer:they are a renewable source

Explanation: because they come from the ocean and the ocean is 75% of the world so there will always be waves.

A string of fifty 15 ohm Christmas tree lights are connected parallelly. When one of them burns , the rest stay lit. The total resistance is :

Total resistance = 3.266 ohm

A break in any one of the path does not interrupt the flow of current in the remaining paths. Sum of the reciprocals of individual resistance is equal to the the reciprocal of the total resistance .

If the two resistances connected in parallel are equal and of the same value, then the total resistance, RT is equal to half the value of one resistor.

A string of fifty 15 ohm Christmas tree lights are connected parallelly. One of them burns, the rest stay lit,

Total resistance in parallel = 1/R1 +1/R2+....

= 1/15 +1/15+...…. 1/15  (49 times as one burns )

= 1/15* 49

= 3.266 ohm.

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Answer:

4,524,660 N

Explanation:

Assuming the submarine's density is uniform, 1/9th of the submarine's mass is equal to the mass of the displaced water.

m/9 = (1026 kg/m³) (50 m³)

m = 461,700 kg

mg = 4,524,660 N

The radioactive sample has a half-life of approximately 83.5 hours. The closest match among the supplied options is (c) 30.1 hours, but none of them exactly match.

The formula for half-life, t12 = ln(2) /, can be used to address this issue. Ln(2) is the natural logarithm of 2, which is roughly 0.693.

We can use the following formula to determine the decay constant: N = N0 * e^(-λt)

By solving for it, we can use this formula to determine the decay constant (): λ = ln(N0/N) / t

With the values provided, we have: t = 3.73 days, N0 = 0.892 g, N = 0.114 g.

Changing the minutes to hours: 3.73 days times 24 hours every day is 89.52 hours.

The equation for is changed to read as follows: = ln(0.892 g/0.114 g) / 89.52 hours 0.0083 per hour. t12 = ln(2) / / 83.5 hours is the equation for half-life after adding the value for. The radioactive sample's half-life is therefore roughly 83.5 hours.

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Answer:

2 hrs

Explanation:

time = distance ÷ speed

Answer:

Advantages. It is suitable to measure low temperatures because of its freezing point is low. It is less toxic than mercury thermometer and hence high on safety factor for human and environment. It has a greater value of temperature coefficient of expansion than a mercury thermometer.

Answer:

1. The stone will strike the ground 49.46 m from the base of the cliff

2. A) Approximately 0.542 seconds

B)  Approximately 3.69 m/s

3. A) The time the ball spends in the air is approximately 4.0775 s

B) The horizontal range is approximately  141.25 m.

Explanation:

1. The time it takes the stone to land is given by the equation, t = √(h/(1/2 × g)

∴ t = √(30/(1/2 × 9.81)) ≈ 2.473 seconds

The horizontal distance covered by the stone in that time = 20 × 2.473 ≈ 49.46 m

The stone will strike the ground 49.46 m from the base of the cliff

2. A) The time the ball spends in the air = t = √(h/(1/2 × g)

∴ t = √(1.44/(1/2 × 9.81)) ≈ 0.542 seconds

B) The initial horizontal velocity, u = Horizontal distance/(Time) = 2/0.542 ≈ 3.69 m/s

The initial horizontal velocity ≈ 3.69 m/s

3. A) The time the ball spends in the air is given by the following equation;

t = 2 × u × sin(θ)/g = 2 × 40 × sin(30)/9.81 ≈ 4.0775 s

t ≈ 4.0775 s

B) The horizontal range, R, of the  ball is given by the equation for the range of a projectile as follows;

\(Range, R = \dfrac{u^2 \times sin (2 \cdot \theta) }{g}\)

Substituting the known values, gives;

\(Range, R = \dfrac{40^2 \times sin (2 \times 30^{\circ}) }{9.81} \approx 141.25 \ m\)

The horizontal range ≈ 141.25 m.

Force of air and gravitational force are acting on the ball when launched from the slingshot.

When a ball is launched from a slingshot, during its travel force of air as well as gravitational force are acting on the ball because there is air which moves opposite to the motion of a ball whereas the force of gravity attracts the ball downward.

The force of air slows down the motion of the ball while on the other hand, the force of gravity brings the ball to the ground due to attraction. If these two forces are not present then the ball continues its motion forever so we can conclude that force of air and force of gravity are acting on the ball when launched from the slingshot.

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Answer:

75000J

Explanation:

Given parameters:

Force applied on the car  = 1000N

Distance moved  = 75m

Unknown:

Work done  = ?

Solution:

The work done by a body is the force applied to move it through a particular distance.

So;

    Work done  = Force x distance

Therefore;

    Work done  = 1000 x 75  = 75000J

Answer is A 2.8 x 10^-8 N, to the right

The net gravitational force on mass A due to masses B and C is required.

The correct option is a. \(2.8\times 10^{-8}\ \text{N}\) to the right.

G = Gravitational constant = \(6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2\)

\(r_{AB}\) = Distance between A and B = 0.4+0.1 = 0.5 m

\(r_{AC}\) = Distance between A and C = 0.1 m

\(m_A=m_B=m_c\) = Mass of each particle = 2 kg

The required force is

\(F_A=F_{AB}+F_{AC}\\ =\dfrac{Gm_Am_B}{r_{AB}^2}+\dfrac{Gm_Am_C}{r_{AC}^2}\\ =Gm^2\left(\dfrac{1}{r_{AB}^2}+\dfrac{1}{r_{AC}^2}\right)\\ =6.674\times 10^{-11}\times 2^2\left(\dfrac{1}{0.5^2}+\dfrac{1}{0.1^2}\right)\\ =2.8\times 10^{-8}\ \text{N}\)

The magnitude of force will be \(2.8\times 10^{-8}\ \text{N}\)

The direction will be towards the right since C is closer to A.

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Answer:

The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)

Answer:

The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)

Explanation:

The net work done on a particle equals the change in the particle's kinetic energy:

Complete question:

A hydraulic jack is used to lift a car by applying a force of 120N at the pump piston, if the area of the ram and pump piston are 100cm squared and 1m squared respectively. What is the weight of the car?

Answer:

the weight of the car is 1.2 N

Explanation:

Given;

applied force, F₁ = 120 N

area of the effort (pump), A₁ = 1 m²

area of the load (ram), A₂ = 100 cm² = 1 x 10⁻² m²

let the weight of the car = F₂

The applied pressure is constant and the following equations can be used to calculate the weight of the car.

\(P = \frac{F_1}{A_1} = \frac{F_2}{A_2} \\\\F_2 = \frac{F_1 \times A_2}{A_1} \\\\F_2 = \frac{120 \times (1\times 10^{-2})}{1} \\\\F_2 = 1.2 \ N\)

Therefore, the weight of the car is 1.2 N

Answer:

(c)Tension = 162 N

(d) Increase

Explanation:

(c)

In order for the cable to lift the plate at uniform velocity, the tension in the cable must be equal to the weight of the plate.

\(Tension = Weight = (Density)(Volume)\\Tension = (2.7\ x 10^3\ kg/m^3)(1\ m)(2\ m)(0.03\ m)\\\)

Tension = 162 N

(d)

The tension with upward acceleration will become:

\(Tension = m(g+a)\\Tension = mg + ma\\Tension = Weight + ma\\\)

Hence, the tension will increase by an amount equal to "ma".

A box is being dragged across the floor at a constant speed by a rope pulling horizontally on it. friction is not negligible. Weight (W), the normal force( N), and the tension force( T )in the string will all be forces operating on the box.

The force transmitted through a rope, string, or wire while opposing forces pull it is referred to as the tension force.

Energy is drawn equally from both ends by the tension force, which is exerted along the entire length of the wire.

A box is being dragged across the floor at a constant speed by a horizontally pulling rope. Friction has a big impact.

As a result, the weight W, normal force N, and tension force T in the string will all be acting on the box.

Your question is incomplete but most probably your full question was

A box is being dragged across the floor at a constant speed by a rope pulling horizontally on it. friction is not negligible. identify all the forces acting on the box.

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The following will happen if we looked at a smooth tabletop with a powerful magnifier:

A thin layer of moisture on the surface makes it somewhat 'sticky'

Hence, option (a) is the correct choice.

If you looked at a smooth tabletop with a powerful magnifier, you would likely see very small bumps and grooves on the surface.

These small imperfections in the surface can cause a frictional force to be produced when a block slides across it.

The frictional force is due to the interaction between the small bumps and grooves in the surface and the surface of the block.

The bumps and grooves can create small points of contact between the two surfaces, which resist the motion of the block and produce a frictional force.

This is similar to the way that the sticky notes in the analogy produce a frictional force by sticking to each other.

The small imperfections in the surface can be thought of as the 'stickiness' in the analogy.

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Answer:

A is a solid. C is a gas. In solid an liquid the particals are touching. In C, the particals have less affect on each other because they are so far apart.

Answer:

easy  how you would die from a plastic bag being over your head is by sufacation the oxegen you breath with soon become carbon dioxide and kill you is it gets to your cellular reperation atory

Explanation:

Answer:

Explanation:

At t = 1 velocity = 0

At t = 3 velocity = 3

slope of the line = 3-0 / 3-1 = 3/2

At t = 2

velocity = 3/2 x ( 2 - 1 )

= 1.5 m /s

velocity at t = 2 is 1.5 m /s

Position at t = 2 :

displacement at t = 2

area of graph upto  t = 2

= 1 / 2 x 1 x 1.5 = .75

position at t = 2 :

= initial position + displacement

= 10 + .75 = 10.75 m

position at 6 s :

displacement at t = 6

area of curve upto  t = 6

= 1 / 2 x 2 x 3 +  3 x 3 + 1/2 x 3 x ( 4.5 - 3 )

= 3 + 9 + 2.25

= 14.25 m

position at t = 6

= initial position + displacement

= 10 + 14.25 = 24.25 m

position at 9 s :

displacement at t = 9

area of curve upto  t = 9

= 1 / 2 x 2 x 3 +  4 x 3 + 1/2 x 4 x ( 5 - 3 )- 1/2 x 2 x 1.5

= 3 + 12 + 4 - 1.5

= 17.5  m

position at t = 9

= initial position + displacement

= 10 + 17.5 = 27.5  m

Answer: Its the last one: The position of the block will change

Explanation: Since the forces are unbalanced, one side of the block will be stronger, therefore making the block move.

Answer:portion of the block will change

Explanation:

Answer:  Fluorine (F) and chlorine (Cl) are two good choices.  See reasoning below.

Explanation:  Mendeleev arranged the elements in the periodic table in a manner that each group, or vertical column, contained elements that exhibited similar chemical properities.  He even reserved spaces for undiscovered elements when none of the known elements had properties similar to those above or below the known elements in a group.  An interesting event illustrates this property.  In December 2010, NASA announced an important new discovery that was to be broadcast on live TV, creating much speculation (e.g., UFO's are real and we'll see the aliens sequestered at Roswell!!!  I knew it.).  

The actual announcement:

"Dec. 2, 2010:  NASA-supported researchers have discovered the first known microorganism on Earth able to thrive and reproduce using the toxic chemical arsenic. The microorganism, which lives in California's Mono Lake, substitutes arsenic for phosphorus in the backbone of its DNA and other cellular components."

A look at the periodic table shows arsenic just below phosphorus, in the same group.   This was an unexpected find, but it might have been predicted with an understanding of how the elements are organized.  

Side note:  look how close silicon is to carbon.  Can there be a life form whose being is made up of silicon instead of carbon?  Sure . . . why not?  Maybe not on Earth, but . . . wait . . . did that marble statue just wink?

Answer:

0.048J/g°C

Explanation:

Given parameters:

Mass of Mercury  = 25g

Initial temperature  = 25°C

Final temperature  = 155°C

Amount of heat absorbed  = 455J

Unknown:

Specific heat capacity of mercury  = ?

Solution:

To solve this problem, we use the expression below:

      Q  = m x C  x Δt

Q is the heat absorbed

m is the mass

C is the unknown specific heat capacity

Δt is the change in temperature;

          455  = 25 x C x (155  - 25)

          455  = 25 x C x 130

             C  = 0.048J/g°C

\(\text{Weight,}~ \\\\W = mg = 45 \times 9.8 = 441~ N\\\\\text{The weight of the object on earth is 441 N}.\)

No, changing the hanging mass would not affect the values obtained for the moment of inertia of the ring and disk as they are intrinsic properties of the objects.

If the experiment is repeated with a larger hanging mass, it would not change the values obtained for the moment of inertia of the ring and disk. The moment of inertia is an intrinsic property of an object and does not depend on external factors such as the hanging mass.

The moment of inertia of an object depends on its mass distribution and how the mass is distributed around the axis of rotation. In the case of a ring and a disk, their moments of inertia can be calculated using their respective formulas.

The moment of inertia of a ring is given by I = MR², where M is the mass of the ring and R is its radius. The moment of inertia of a disk is given by I = (1/2)MR², where M is the mass of the disk and R is its radius. These formulas are derived based on the mass distribution and geometry of the objects.

During the experiment, the hanging mass is used to apply a torque to the rotating system, causing it to accelerate. By measuring the angular acceleration and knowing the torque applied, we can use the rotational dynamics equation τ = Iα to solve for the moment of inertia (I) of the rotating object. The equation relates the torque (τ), moment of inertia (I), and angular acceleration (α).

In this experiment, the moment of inertia of the ring was calculated using the known torque and angular acceleration, allowing for the determination of its experimental value. The moment of inertia of the ring was calculated in this manner because the rotational dynamics equation provides a direct relationship between torque, angular acceleration, and moment of inertia.

Therefore, changing the hanging mass used in the experiment would not affect the values obtained for the moment of inertia of the ring and disk since the hanging mass does not influence their intrinsic mass distributions or geometries.

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