Let’s discuss some particle physics. In 1897, physicist J.J. Thomson conducted a seminalexperiment with a cathode ray tube. In the experiment, a beam of an unknown particle of chargeqandmassmis shot through the tube onto a fluorescent screen, at some speedvand in thex-direction. Uponimpact the beam leaves a bright spot on the screen. Sincevis very fast, effects of gravity are negligible andthe beam forms essentially a straight line normal to the screen.

To eliminate this deflection, an external uniform magnetic field of magnitudeBis turned on as well. What should be the direction and magnitudeBof this field such that ∆z= 0?

Answer:

the impulse imparted to the ball is 4.158 kg.m/s

Explanation:

Please find attached image for diagram.

Given;

initial speed of the ball from point A, u = 54km/h = 15 m/s

angle of deflection of the ball along BOP, θ = 45⁰/2 = 22.5⁰

mass of the ball, m = 0.15 kg

the final speed of the ball along B,  = u (in reverse direction)

The initial momentum of the ball, P₁ = mucosθ

The final momentum of the ball is P₂ in reverse direction to P₁.

the impulse imparted to the ball = change in momentum of the ball

J = ΔP = P₂ - P₁

J = mucosθ - (-mucosθ)

J = mucosθ + mucosθ

J = 2mucosθ

J = 2(0.15)15cos(22.5⁰)

J = 4.158 kg.m/s

Thus, the impulse imparted to the ball is 4.158 kg.m/s

It should be noted that the reason for the difference in brightness is that the resistance is lower on the right than on the left.

From the complete information, the resistance that's given in the first two LED circuits is R. Also, the resistance that is given in the two LED circuits onto the right is R/2.

From Ohm's law, I = V/R. This implies that I is indirectly proportional to R. Therefore when the resistance is decreased, there will be an increase in current flow.

Therefore, the glow of the two LEDs on the right is more than those on the left because the resistance is lower on the right than on the left.

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1) Robert Brown, a Scottish botanist who was born on December 21, 1773, is best known for his descriptions of cell nuclei and what is now known as Brownian motion—the continuous motion of tiny particles in solution.

2) Robert Brown found no cell. His discovery of Brownian motion—the haphazard movement of small particles in a surrounding solution—made him most famous. He also created a different classification system for plants.

3) Robert Brown discovered the existence of a structure within the cells of orchids and many other plants in 1831 while researching the fertilization processes of plants in the Orchidaceae and Asclepiadaceae families. He named this structure the "nucleus" of the cell.

4) Hooke observed cork under his microscope and noticed tiny box-like cavities, which he depicted and referred to as cells. He had stumbled across plant cells! Hooke's finding laid the groundwork for cell theory by revealing cells as the tiniest constituents of life.

5) Emmett Brown was an actor best remembered for his roles in St. Elmo's Fire, The Outsiders, and Rumble Fish from 1983. (1985). He passed away on April 12, 1989.

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Answer:

see explanation

Explanation:

beam balance

temperature degree celsius

newton meter per second

weight.

hope I was helpful to tou

Answer:

The forward motion of a body in space, such as a planet or moon, and the pull of gravity on it from another body in space.

Explanation:

Earth and many other bodies—including asteroids, comets, and the other planets—move around the sun in curved paths called orbits. Generally, the orbits are elliptical, or oval, in shape. Because of the sun’s relatively strong gravity, Earth and the other bodies constantly fall toward the sun, but they stay far enough away from the sun because of their forward velocity to fall around the sun instead of into it. As a result, they keep orbiting the sun and never crash to its surface. The motion of Earth and the other bodies around the sun is called orbital motion. Orbital motion occurs whenever an object is moving forward and at the same time is pulled by gravity toward another object.

Answer:

Conduction: Touching a stove and being burned. Ice cooling down your hand. .

Explanation:

Answer:

2.0 m/s

Explanation:

Remember that F = maF=maF, equals, m, a, where FFF is the net force, mmm is the mass of the object, and aaa is its acceleration. To solve this problem, we will need to first find the net force acting on the piano and then use F = maF=maF, equals, m, a to find the piano’s acceleration.

Hint #22 / 4

The net force is the sum of all of the forces on the piano. In this circumstance, Elvis's and Noah's forces are in the same direction, and the friction force is in the opposite direction. So, our net force equation would be:

​

=400 N+150 N−100 N

=450 N

Now, we can use F = maF=maF, equals, m, a and solve for the acceleration:

225 kg

450 N

=2.0

The acceleration of the piano is 2.0 m/s

The acceleration of the piano is 2m/s².

We know the formula for force is

F =ma

where f is force, m is mass and a is acceleration.

We need to calculate the total force used in moving the piano

F= Total force- Frictional force

F= 400N+150N-100N= 450N.

We have F=450N and m= 225kg

F=ma

450N= 225ˣ a

a= 450/225

a= 2m/s²

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In the parallel combination, the current through R3 is 3 A. In the series combination, the current through R3 is 0.75 A.

To determine the current through resistor R3 in both the parallel and series combinations, we need to apply Ohm's Law and the appropriate formulas for calculating total resistance and current in each configuration.

First, let's consider the parallel combination:

In a parallel combination, the voltage across each resistor is the same. Therefore, the voltage across R3 is also 90 V.

Using Ohm's Law (V = I × R), we can calculate the current flowing through R3 in the parallel combination:

I_parallel = V / R3

= 90 V / 30 Ω

= 3 A

So, in the parallel combination, the current through R3 is 3 A.

Now, let's consider the series combination:

In a series combination, the total resistance is the sum of the individual resistances:

R_total = R1 + R2 + R3

= 60 Ω + 30 Ω + 30 Ω

= 120 Ω

To find the current through the series combination, we can use Ohm's Law:

I_series = V / R_total

= 90 V / 120 Ω

= 0.75 A

Therefore, in the series combination, the current through R3 is 0.75 A.

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Note the complete questions is User

90 V R₁=60 R2=  30, R3 = 30

Based on the circuit above, what would be the current through the R3 resistor in parallel and in series combinantion.

Answer:

The atomic and molecular interactions unveil the bulk properties of matter in our environment by ways of the fact that everything in the whole universe is made of either atoms, molecules or even ions

How matter is made up of atoms and molecules?

It has been proven practically everything in the whole universe is matter and everything which interact with matter is also matter. This explains to us the reasons why matter could be atoms, molecules or ions.

That being said, some substances (matter) is made up of atoms of elements, some made up of molecules or atoms and molecules and others ions or both. However, matter is anything which has mass and occupies space.

In conclusion, we can now conclude from the explanation above that the properties of matter are as a result of the interaction which exists between matter at the atomic and molecular level.

Answer:

Hi

The acceleration in m/s 2 is:

a= (v-u)/t

a= (29.0-0)/5.75

a= 5.043478261

a= 5.04 (3s.f)

Work is defined as the process of energy transfer to the motion of an object through the application of force. Power is defined as the amount of energy transferred in unit time.

Energy is that which has the ability to cause change in matter.

Energy is the quantitative property that is transferred to a body or to a physical system, recognizable in the performance of work and in the form of heat and light. Energy is a conserved quantity—the law of conservation of energy states that energy can be converted in form, but not created or destroyed.

So in simple definition we can say that energy is that which has the ability to cause change in matter.

Based on the given statements we can classify them as;

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Answer:

\(1.6 \,\,\,10^{-4}\)

Explanation:

Recall that scientific notation involves expressing the number in powers of ten, and using a multiplicative factor whose value is larger than or equal to 1 (one) and strictly smaller than 10 in the form:

\(a\,\,\,10^n\\with\,\,\,1\leq a<10\)

and with n an integer value.

in our case, notice that 0.00016 = 16 / 100000 = 1.6 / 10000

and that denominator can be expressed as: 10^4, and can also be brought to the numerator as a negative power of ten:

1.6 / 10000 = 1.6 / 10^4 = 1.6 * 10^(-4)

\(\frac{16}{100000} =\frac{1.6}{10000}=\frac{1.6}{10^4}=1.6\,\,\,10^{-4}\)

Sound travel faster in metal than air becoz molecules of metal i.e solid are closely packed and it will take less time for sound waves to travel from one molecule to another whereas in air i.e in gas,molecules are far apart so it will take more time for sound waves to travel.In case of vacuum there are no molecules at all so sound is not there.that is why in space we cannot hear anything.

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Answer:

it is safe to stand at the end of the table

Explanation:

For this exercise we use the rotational equilibrium condition

         Στ = 0

         W x₁ - w x₂ - w_table x₃ = 0

         M x₁ - m x₂ - m_table x₃ = 0

where the mass of the large rock is M = 380 kg and its distance to the pivot point x₁ = 850 cm = 0.85m

the mass of the man is 62 kg and the distance

            x₂ = 4.5 - 0.85

            x₂ = 3.65 m

the mass of the table (m_table = 22 kg) is at its geometric center

            x_{cm} = L/2 = 2.25 m

            x₃ = 2.25 -0.85

            x₃ = 1.4 m

let's look for the maximum mass of man

            m_{maximum} = \(\frac{ M x_1 -m_{table} x_3}{ x_2}\)

let's calculate

             m_{maximum} = \(\frac{ 380 \ 0.85 - 22 \ 1.4}{3.65}\)(380 0.85 - 22 1.4) / 3.65

             m_{maximum} = 80 kg

we can see that the maximum mass that the board supports without turning is greater than the mass of man

             m_{maximum}> m

consequently it is safe to stand at the end of the table

The velocity of the particle is given by the derivative of the position vector:

\(\vec v = \dfrac{\mathrm d\vec r}{\mathrm dt} = (2ct-6dt^2)\,\vec\imath + (4ct-2dt)\,\vec\jmath\)

(a) The particle is moving in the x-direction when the y-component of velocity is zero:

\(4ct-2dt = 2t (2c - d) = 0 \implies t=0\)

But we want t > 0, so this never happens, unless 2c = d is given, in which case the y-component is always zero.

(b) Similarly, the particle moves in the y-direction when the x-component vanishes:

\(2ct-6dt^2 = 2t (c - 3dt) = 0 \implies t=0 \text{ or }c-3dt = 0\)

We drop the zero solution, and we're left with

\(c-3dt = 0 \implies c=3dt \implies \boxed{t = \dfrac c{3d}}\)

In the case of 2c = d, this times reduces to t = c/(6c) = 1/6.

The electrostatic interactive force between electrical charges is exactly the same as the gravitational interaction between them is referred to as the incorrect statement and is denoted as option A.

This is defined as the force of attraction or repulsion which exists in a charged body and it is different from the gravitational force interaction between two bodies.

This is because gravitational force is the force of objects as a result of its mass while  electrostatic force is the force of an object due to charge which is therefore the reason why option A was chosen.

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The mass of the planet orbiting through the path can be calculated using its period of orbit, and radius . Here, the mass of the planet be 7.11 × 10³⁰ kg.

The orbital period of a spatial object is the time taken to revolve around a star or another spatial object through out an elliptical orbit.

The mass of the star can be calculated using the formula :

m = 4 π r³/G T²

given,

r = 4 Au = 4 × 1.5 × 10¹¹ m

T = 4 years = 12.6 × 10⁷ seconds.

G = gravitational constant 6.67 × 10⁻¹¹ m³/kg s²

Then M = 4 × 3.14 × (4 × 1.5 × 10¹¹ m)³/ 6.67 × 10⁻¹¹ m³/kg s² × ( 12.6 × 10⁷ s)²

             = 7.11 × 10³⁰ kg.

Therefore, the mass of the planet is 7.11 × 10³⁰ kg..

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Answer:

C. That the small portion of the space we can see is truly representative of all the rest of the universe that we cannot see.

Explanation:

The Cosmological Principle assumes that the small portion of the universe that we can see is representative of the entire universe, even though we can only directly observe a tiny fraction of it. It's an assumption used by Cosmologists to simplify their models of the universe.

Answer:

700km

Explanation:

going was 250km and leaving was 450

250+450=700

( a ) 966 Ps is expressed in SI unit as 9.66 x 10⁻¹⁰ s

( b ) 966 fs is expressed in SI unit as 9.66 x 10⁻¹³ s

( c ) 48 ns is expressed in SI unit as 4.8 x 10⁻⁸ s

( d ) 640  μsis expressed in SI unit as 6.4 x 10⁻⁴ s

A standard way of expressing a derived quantity or fundamental quantity in terms of the SI unit is always in power of 10.

The standard form of the given quantity is calculated as follows;

( a ) 966 Ps = 966 x 10⁻¹² s = 9.66 x 10⁻¹⁰ s

( b ) 966 fs =  966 x 10⁻¹⁵ s = 9.66 x 10⁻¹³ s

( c ) 48 ns = 48 x 10⁻⁹ s = 4.8 x 10⁻⁸ s

( d) 640  μs = 640 x 10⁻⁶ s = 6.4 x 10⁻⁴ s

Thus, the standard form of a quantity in SI unit involves the power of 10.

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The time will be the same for both horizontal and vertical component. The initial speed is 10.7 m/s

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

Given that a person standing at the edge of a seaside cliff kicks a rock horizontally of the cliff from a height of 52 m and it lands a distance of 35 m from the base of the cliff.

The rock will move vertically downward with initial velocity = 0. The time taken will be constant. That is, same horizontally.

Let us first calculate the time by using the formula

h = ut + 1/2gt²

Where

Substitute all the necessary parameters into the formula

52 = 0 + 1/2 × 9.8 × t²

52 = 4.9t²

t² = 52/4.9

t² = 10.6

t = √10.6

t = 3.26 s

The speed at which the rock was initially kicked can be found by

R = Ut

35 = U × 3.26

U = 35/3.26

U = 10.7 m/s

Therefore, rock was initially kicked at a speed of 10.7 m/s

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Given:

The mass of the block, m=4 kg

The initial height of the block, h=5 m

The spring constant of the spring, k=120 N/m

To find:

A. Initial potential energy of the block.

B. Velocity of the block at the bottom of the incline.

C. The compression in the spring.

Explanation:

A.

The gravitational potential energy is the energy possessed by an object due to its height.

Thus the initial potential energy of the block is given by,

Where g is the acceleration due to gravity.

On substituting the known values,

B.

From the law of conservation of energy, the total energy of an isolated system always remains constant. As the block slides down the apparatus, it loses its potential energy. And as there is no friction, the potential energy lost will be converted completely into the kinetic energy of the block.

Thus, when the block is at the bottom of the incline,

Where v is the velocity of the block when it is at the bottom of the incline.

On substituting the knwon values,

C.

From the law of conservation of energy, the total energy of the system remains the same.

Thus, when the block is stopped, the kinetic energy of the block is completely converted into the elastic potential energy of the spring. Thus the initial potential energy of the block is equal to the spring potential energy stored in the spring when it is compressed and the block is stopped.

Thus,

Where x is the compression in the spring.

On substituting the known values,

Final answer:

A. The initial potential energy of the block is 196 m

B. The velocity of the block at the bottom of the incline is 10 m/s

C. The compression in the spring 1.81 m

The right sphere will acquire an equal and opposite net positive charge to balance the negative charge on the left sphere.

Since the left sphere is attracted to the positively charged rod, it means that the left sphere acquires a temporary negative charge due to induction.

The positive charge on the rod repels electrons in the left sphere, causing them to move away from the rod side and accumulate on the opposite side, resulting in a net negative charge on the left sphere.

According to the principle of charge conservation, the net charge on the system must remain zero. Therefore, the right sphere acquires an equal and opposite net positive charge to balance the negative charge on the left sphere.

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When you draw all 4 of those moves, you see that he winds up 9m North and 11m West of where he started.

So the magnitude of his displacement is

D = √(9² + 11²)  m²

D = √(81 + 121)  m²

D = √(202)  m²

D = 14.2 m

The direction is arctan(-11/9) = 50.7° west of North

Answer:

liquid

Explanation:

In compressing the spring by 0.5 m, Cho stores

1/2 (25 N/m) (0.5 m)² = 3.125 J

of potential energy. As the spring relaxes, some of this potential energy is converted into kinetic energy K in the box and the rest is lost to heat H due to friction:

K - H = 3.125 J

By the work-energy theorem, the total work W performed on the box during this process is equal to the change in the box's kinetic energy. It starts at rest when the spring is compressed, so

W = K - 0 = 1/2 (1.5 kg) v²

By Newton's second law,

• the net vertical force on the box is

∑ F (ver) = n - mg = 0

where n is the magnitude of the normal force from contact with the table, and mg is the weight of the box. It follows that

n = mg = (1.5 kg) (9.8 m/s²) = 14.7 N

• the net horizontal force is

∑ F (hor) = r - f = ma

where r is the mag. of the restoring force, f = 0.2 n is the mag. of kinetic friction, and a is the resulting acceleration of the box. Friction exerts an opposing force of

f = 0.2 (14.7 N) = 2.94 N

The work done by the spring as it relaxes is 3.125 J. The work done by friction is (2.94 N) (0.5 m) = 1.47 J. Therefore the total work done on the box is

W = 3.125 J - 1.47 J = 1.655 J

Find the speed of the box v when it returns to the equilibrium position:

1/2 (1.5 kg) v² = 1.655 J

v² = (3.31 J) / (1.5 kg)

v ≈ 1.49 m/s

I assume the rest of the table is not scarred. (If it is, the box would come to a stop well before 2.30 seconds have passed, or before the box can slide for 2.30 m.) Then the box leaves the table with speed v at an angle of 75°, so that the horizontal distance x that it covers in the air after t seconds is

x = v cos(75°) t

and its tolerable height y is

y = 2.5 m + v sin(75°) t - g/2 t²

(if y turns out negative for some t, that means it has fallen more than 2.5 m and cannot land safely)

The pile is 3.5 m away from the table, so with its launch speed the box travels this distance in time t such that

v cos(75°) t = 3.5 m

t = (3.5 m) / ((1.49 m/s) cos(75°)) ≈ 9.1 s

After such time, the box would have height

2.5 m + v sin(75°) (9.1 s) - g/2 (9.1 s)² ≈ -390.51 m

i.e. it would be nearly 400 m underground! So no, the box will not land safely.