Determine the applied force required to accelerate a 2.25 kg object rightward with a
constant acceleration of 1.50 m/s2 if the force of friction opposing the motion is 18.2 N.
(Neglect air resistance.)

Answer:

If the tunnel is straight, it will have 2 hours to pass through it

Explanation:

Train speed distance time

1km/hr. 1km. 1hr. 1km/hr. 2km. 2hr

Answer:

The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

Explanation:

Given;

initial temperature of the liquid, t₁ = 76.3  +/-  0.4⁰C

final temperature of the liquid, t₂ = 67.7  +/-  0.3⁰C

The change in temperature of the liquid is calculated as;

Δt = t₂  -  t₁

Δt = (67.7 - 76.3)  +/-  (0.3 - 0.4)

Δt = (-8.6)  +/-  (-0.1)

Δt = 8.6 +/- 0.1 ⁰C

Therefore, the fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

Answer:

Explanation:

1 mole of gas contains 6.02x10^23 molecules. That is the Avogadro's number. To find the number of moles, you divide the number of molecules by the Avogadro's number.

The correct answer is C.

Answer:

Explanation:

ans is C. divided by Avogadro no.

Answer:

From Ohms Law

V=IR

V=5volts

R=33.0hms

I=V/R

I= 5/330

=0.0152A

The current flowing through the cross-section of the resistor is 0.0152A.

Given the data in the question;

Ohm's law states that "the current flowing through the cross-section of a conductor is directly proportional to the voltage across it".

It is expressed as;

[tex]V = IR[/tex]

Where V is voltage, I is current and R is the Resistance.

To determine the current flowing through the resistor, we substitute our given values into the expression above.

[tex]V = IR\\\\I = \frac{V}{R}\\ \\I = \frac{5.00V}{330\ Ohm} \\\\I = 0.0152A[/tex]

Therefore, the current flowing through the cross-section of the resistor is 0.0152A.

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Answer:

a)   w = 873 rev,  b)  θ = 97.75 rev

Explanation:

This is a rotation kinematics exercise

         w = w₀ + α t

         θ = θ₀ + w₀ t + ½ α t²

let's start by reducing the magnitudes to the SI system

         w₀ = 300 rpm (2pi rad / 1 rev) (1 min / 60s) = 31.42 rad / s

          α = 6 rad / s²

a) let's look for the angular velocity

            w = 31.42 + 6 10

             w = 91.42 rad / s

b) θ₀ = 0

             θ = 0 + 31.42 to + ½ 6 10²

             θ = 614.2 rad

As they ask for the result in rpm and revolutions, let's carry out the reduction

         w = 91.42 rad / s (1 rev / 2pi rad) (60 s / 1min)

         w = 873 rev

         θ = 614.2 rad (1 rev / 2pi rad)

         θ = 97.75 rev

Answer:

An orbit is a regular, repeating path that one object in space takes around another one. An object in an orbit is called a satellite. A satellite can be natural, like Earth or the moon. ...Planets, comets, asteroids and other objects in the solar system orbit the sun

Explanation:

Answer:

a) The closest object Anne can see, while wearing Bill's glasses is at approximately 22.39 cm relative to her eyes

b) The closest object Bill can see while wearing Anne's glasses is at approximately 26.38 cm relative to his eyes

Explanation:

The point where Bill has a near point = 125 cm

The point where Anne has a near point = 75.0 cm

Their vision are both corrected to the normal near point = 25.0 cm

The distance of their glasses from the eye, d = 2.0 cm

The lens formula is presented as follows;

[tex]\dfrac{1}{f} =\dfrac{1}{d_0} +\dfrac{1}{d_i}[/tex]

The required distance of the of the object from Bill's glass, [tex]d_{oB}[/tex] = 25 cm - 2.0 cm = 23.0 cm

The distance of the of the image from Bill's glass, [tex]d_{iB}[/tex] = -(125 cm - 2.0 cm) = -123 cm

Therefore, for Bill, we have;

[tex]\dfrac{1}{f_B} =\dfrac{1}{d_{0B}} +\dfrac{1}{d_{iB}}[/tex]

Plugging in the values gives;

[tex]\dfrac{1}{f_B} =\dfrac{1}{23} -\dfrac{1}{123} = \dfrac{100}{2,829}[/tex]

Therefore;

[tex]f_B[/tex] = 2,829/100 cm = 28.29 cm

The required distance of the of the object from Anne's glass, [tex]d_{oA}[/tex] = 25 cm - 2.0 cm = 23.0 cm

The distance of the of the image from Anne's glass, [tex]d_{iA}[/tex] = -(75 cm - 2 cm) = -73 cm

The focal length for Anne is therefore;

[tex]\dfrac{1}{f_A} =\dfrac{1}{d_{0A}} +\dfrac{1}{d_{iA}}[/tex]

Plugging in the values gives;

[tex]\dfrac{1}{f_A} =\dfrac{1}{23} -\dfrac{1}{73} = \dfrac{50}{1,679}[/tex]

Therefore, we have;

[tex]f_A[/tex] = 1,679/50 cm = 33.58 cm

a) When Anne wears Bill's lasses, we have;

[tex]\dfrac{1}{f_B} =\dfrac{1}{d_{0A}} +\dfrac{1}{d_{iA}}[/tex]

Therefore, we get;

[tex]\dfrac{1}{28.29} =\dfrac{1}{d_{0A}} -\dfrac{1}{73}[/tex]

[tex]\dfrac{1}{d_{0A}} = \dfrac{1}{28.29} + \dfrac{1}{73} = \dfrac{10,129}{206,517}[/tex]

[tex]d_{0A}[/tex] ≈ 20.39 cm

The distance of the closest object Anne can see, from her eye, [tex]d_{oe}[/tex] =  [tex]d_{0A}[/tex] + d

∴ [tex]d_{oe}[/tex] ≈ 20.39 cm + 2.0 cm = 22.39 cm

The distance of the closest object Anne can see, from her eye, [tex]d_{oe}[/tex] ≈ 22.39 cm

b) The closest object that can be seen when Bill wears Anne's glasses, we have;

[tex]\dfrac{1}{f_A} =\dfrac{1}{d_{0B}} +\dfrac{1}{d_{iB}}[/tex]

Therefore;

[tex]\dfrac{1}{33.58} =\dfrac{1}{d_{0B}} -\dfrac{1}{123}[/tex]

[tex]\dfrac{1}{d_{0B}} = \dfrac{1}{33.58} +\dfrac{1}{123} = \dfrac{7,829}{206,517}[/tex]

∴ [tex]d_{oB}[/tex] ≈ 26.38 cm

The closest object Bill can see while wearing Anne's glasses,  [tex]d_{oB}[/tex] ≈ 26.38 cm.

Answer:

D. displacement

Explanation:

Vector quantities have both magnitude and direction.

Scalar quantities have only magnitude.

Answer:

It's option 1: they have the same density.

Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what  horizontal force is required to just move the blcok ( take gravity as 10m/s2 )

Answer:

108 N

Explanation:

From the question,

Applying

F' = mgμ................ Equation 1

Where F' = Frictional force = horizontal  force required to just move the block,  m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.

From the question,

Given: m = 18 kg, μ = 0.6, g = 10 m/s²

Substitute these values into equation 1

F' = 18×0.6×10

F' = 108 N

Answer:

30 miles

Explanation:

Answer:

Explanation:

Speed and velocity are not the same thing, as I'm sure you have determined while in physics. To find the average speed, we add the total distance run by Kiran and divide that total number of meters by the number of seconds it took to do this (just because I am going to state the speed and velocity in terms of m/s as opposed to m/min). To find the average velocity, we add the displacements together and divide by the total time. First the average speed:

[tex]s=\frac{150+150}{60}=\frac{300}{60}=5\frac{m}{s}[/tex]

Velocity has direction. If Kiran runs to the end of the road (which we will call positive) and then runs back to the start (that direction will be negative), then his displacement fits into the velocity equation as

[tex]v=\frac{150+(-150)}{60}=0[/tex] That means, even though he ran a lot and I'm sure he was very tired, his average velocity for the trip was 0 m/s. When velocity is 0, remember that that means that the total displacement was 0.

Answer:

b) resistance

Explanation:

since resistance is responsible for heating, if it increases the circuit breaker opens breaking the flow of current to the circuit and thus preventing short circuits.

Answer:

B energy can't be created or destroyed

Answer:

60J

Explanation:

Since the book's weight is 30N and weight = mass * gravity (around 10m/s²):

30 = X * 10 ===>   X = 3.

Using the potential energy formula we have Gravity * Height * Mass:

10 * 2 * 3 = 60J

Answer:

A simple pulley is a wheel with a rope that allows you to pull one end and have it lift whatever is on the other end. A modern, common example of this is a crane, often used in construction.

Explanation:

Answer:

Slow-moving rivers generate extensive floodplains and meanders through erosion and deposition. Stream and river deposition can result in the formation of alluvial fans and deltas. Natural levees may be formed by floodwaters. Caves and sinkholes can arise as a result of groundwater erosion and deposition.

Explanation:

s

The erosion and deposition of sediments in a river creates broad floodplains and meanders.

Deposition occurs when sediment, a combination of soil and rock produced by weathering, is eroded and transported to a new area.

Deposition is the act of depositing silt that has been transported by the wind, water, sea, or ice.

Earthen materials are worn away during erosion, a geological process in which they are moved by wind or water.

The removal of soil, rock, or dissolved material from one area on the Earth's crust and subsequent transport to another region for deposition are known as erosional processes. Erosion differs from weathering, which is a static process.

Given data ,

Let the erosion and deposition of sediments be deposited in a river

Now , Pebbles, sand, mud, and salts that have been dissolved in water can all be used to convey sediment. Afterwards, salts may be left behind by organic action.

Now , floodplain is caused when erosion happens

And , A floodplain is a broad, level or nearly level area of land where the stream flows.

Meandering streams that wander from side to side broaden the plain by eroding it during the formation of the plain. Flooding can occur when stream flows overflow from their channel due to very excessive rainfall or quick snow-melt.

Hence , floodplain and meandering occurs due to erosion and deposition

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Explanation:

(≚ᄌ≚)ℒℴѵℯ❤

Answer:

1.45×10⁻¹⁶ C

Explanation:

Applying,

F = kqq'/r²................ Equation 1

Where F = force acting on the alpha particle, k = coulomb's constant, q = charge on the alpha particle, q' = charge on the atomic nucleus, r = distance of seperation between the alpha particle and the atomic nucleus.

make q' the subject of the equation

q' = Fr²/(kq)................. Equation 2

From the question,

Given: F = 79.8 N, r = 2.29×10⁻¹⁴ m, q = 2e = (2×1.6×10⁻¹⁹) = 3.2×10⁻¹⁹ C, k = 9×10⁹ Nm²/C²

Substitute these values into equation 2

q' = [79.8×(2.29×10⁻¹⁴)²]/(3.2×10⁻¹⁹×9×10⁹)

q' = (4.18×10⁻²⁶)/(2.88×10⁻¹⁰)

q' = 1.45×10⁻¹⁶ C

Answer:

B=8.40x10^-5 T

Explanation:

To find the magnetic field at the new current, first you need to find the distance. To do so, alter the original equation from finding the field to finding the distance.

r (distance) = 4 pi x 10^-7 / 2 pi multiplied by 10.0 / 4.20 x 10^-5

which equals 0.04762

Next, just plug the numbers back into the original equation, taking into account the requested changes.

which equals 8.40x10^-5

Also it was right on Acellus ;)

Hope this helps :)

Answer:

43740 N and attractive force

Explanation:

Given that,

Charge 1, q₁ = -54.0 μC

Charge 2, q₂ = +36.0 μC

The distance between charges, r = 2 cm = 0.02 m

We need to find the force between charges. The formula for the force between charges is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Put all the values,

[tex]F=9\times 10^9\times \dfrac{54\times 10^{-6}\times 36\times 10^{-6}}{(0.02)^2}\\\\F=43740\ N[/tex]

As the charges are opposite, the force between them is attractive. Hence, the required force is 43740 N.

hey I read it do you use there a pacific question you want me to answer or something?

Answer:

I also recently read Call of the Wild for school and it was a great book.

Explanation:

Answer:

10

Explanation:

Answer:

weight = mass×gravity

Answer:

their final velocity is 0.091 m/s

Explanation:

Given;

mass of the first train, m₁ = 138,000 kg

mass of the second train, m₂ = 123,000 kg

initial velocity of the first train, u₁ = 0.288 m/s

initial velocity of the seocnd train, u₂ = -0.131 m/s (opposite direction to the first)

Let their common final velocity after been coupled = v

Apply the principle of conservation of linear momentum;

m₁u₁  +  m₂u₂ = v(m₁  +  m₂)

(138,000 x 0.288)    +    (-0.131 x 123,000)   =   v(138,000 + 123,000)

39,744   -   16,113   =  v(261,000)

23,631 = v(261,000)

v = 23,631 / 261,000

v = 0.091 m/s

Therefore, their final velocity is 0.091 m/s

Answer:

About 4,000 K and 10⁻¹⁷ atm

Explanation:

The Big Bang theory states that the Big Bang which is the origin of the universe was about 13.75 billion years ago, and the temperature a few seconds later was 10³²K

The first element began forming at about 3 minutes after the Big Bang with a temperature of 10⁹ K, the nuclei of simple elements

The nuclei of hydrogen and helium began combine with electrons at a temperature of 3,000 K to 4,000 K to form the first neutral atoms. The pressure of the universe at that stage was 10⁻¹⁷ atmospheres

Answer:

R=V/I

R= 2

Explanation:

R = 10V/5A

R = 2ohms