Magnetism Assessment
Magnet A is a very weak magnet and Magnet B is a very strong magnet. They attract each other. The relative attractive force is
greater from magnet A
greater from magnet B
the same for both magnets
unable to be determined
1
2
3
4
5
6
7
8
9

To find the normal force, we can use the equation: normal force = weight + friction force * cos(incline angle).

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Answer:

  F = 3.6 10⁶ N

Explanation:

The expression for the electric force is

         F =  [tex]k \ \frac{q_1 q_2}{r^2}[/tex]

in this case it indicates that the charge q1 is doubled

       q₁ = 2   3 10⁻³ C

       q₁ = 6 10⁻³ C

let's reduce the magnitudes to the SI system

        q₂ = 6 10⁻³ C

        r = 0.3 m

let's calculate

       F = 9 10⁹ 6 10⁻³ 6 10⁻³ / 0.3²

       F = 3.6 10⁶ N

Answer:

Explanation:

We already know that Force is a vector. Weight being a force, is also a vector quantity. Displacement is distance in a specific direction, hence it is a vector quantity too. ... since energy, work done and time are scalar, Power is a scalar quantity

It would increase................

The radius of the hydrogen-atom Bohr orbit shown in the figure is 5.3 nm.

The path that hypothetical electrons take around the nucleus is known as Bohr's orbit.

These orbits are described by Bohr in his hypothesis of the structure of an atom as energy levels or shells where electrons move in a fixed circle around the nucleus.

These orbits resemble solar system orbits, with the exception that they are attracted by electrical forces rather than gravity. The term "ground state" refers to the amount of energy that an electron typically occupies.

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Answer:

Distance is a scalar quantity while displacement is a vector quantity

Explanation:

A scalar quantity represents only the magnitude and does not give any detail about the direction of the quantity for example distance. Distance can be any length measured in any direction (no specific direction)

However, a vector quality represents both the magnitude and direction. For instance displacement is a vector quantity. If direction is not defined then displacement becomes equal to distance.

Answer:

50 million years

Explanation:

light years is the distance light travels in one year given that the supercluster is 100 million light years across the the distance to the center will be half that amount therefore the answer is 50 million years

Answer:

As a bat increases its forward speed, it should decrease the frequency of the emitted pulses to compensate.

decrease

Explanation:

Decreasing the frequency of the emitted pulse will help the bat reduce its frequency caused by its forward motion.  The forward motion shifts the bat's auditory frequency to a higher frequency; consequently, the bat should adjust downwards the frequency of the emitted pulse so the reflected pulse will be in the correct frequency range.

The boat will arrive 2.5 meter in east far from a point directly north on the other side of the river.

The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction.

A boat has a speed of 20m/s in still waters.

The velocity of river is 5m/s in east.

The width of the river is = 10 meter.

Hence, time taken by the board to cross the river = 10/20 second

= 0.5 second.

In this time, the board will move in east direction   = 5×0.5 meter = 2.5 meter.

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Answer:

800J

Explanation:

W = Fs, Work equals force times displacement

in this case, the force is 400N and the displacement is 2 meters.

The regular SI unit for work is joules

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

[tex]T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}[/tex]

where,

T = Radiated surface temperature

[tex]\lambda_{max}[/tex] = peak wavelength

(a)

here,

[tex]\lambda_{max}[/tex] = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

[tex]T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}[/tex]

T = 2987.6 k

(b)

here,

[tex]\lambda_{max}[/tex] = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

[tex]T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}[/tex]

T = 19986.2 k

Answer:

1078 Joules

Explanation:

The computation of the total kinetic energy of the bicycle is shown below:

Given that

mass of bicycle's frame (m) = 6.6 kg

mass of each wheel (M) = 2.2 kg

radius of each wheel (r) = 0.35 m

And, the linear speed (v) = 14 ms

Now

As we know that

Angular velocity (ω) = v ÷ r

= 140 ÷ .35

= 40 rads

Total kinetic energy = translational kinetic energy + rotational kinetic energy

= (1 ÷2 × m  × v^2) + (2 × 1 ÷ 2×I × ω^2)

= (0.5 × 6.6 × [14]^2) + (M × r2 × ω^2)

= 646.8 + (2.2 × 0.35 × 0.35 × [40]^2)

= 646.8 + 431.2

= 1078 Joules

Answer:

0.5 seconds

Explanation:

Work = Fdcos(theta) = 10*2*cos(0) = 20 J

Power = W/t

40 = 20/t

t = 0.5s

The displacement of the iguana between 3 s and 6 s  is 6.71 meters.

The distance traveled by the iguana between 3 s and 6 s is  8.08 meters.

Distance is the sum of an object's movements, regardless of direction.

The term "displacement" refers to a shift in an object's position.

According to the graph:

The displacement of the iguana between 3 s and 6 s

= √{ (3-6)²+(6-0)²} meters

= 6.71 meters.

The distance traveled by the iguana between 3 s and 6 s

= [ √{ (3-5)²+(6-6)²} +√{ (5-6)²+(6-0)²}] meters

= [2+ 6.08] meters

= 8.08 meters.

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Answer:

70.56 J

Explanation:

The increase in potential energy of the ball can be calculated using the formula:

PE = mgh

where:

PE is the increase in potential energy

m is the mass of the ball (6 kg)

g is the acceleration of gravity (9.8 m/s²)

h is the height the ball is lifted (1.2 m)

Substituting in the values, we get:

PE = (6 kg)(9.8 m/s²)(1.2 m)

This simplifies to:

PE = 70.56 J

So the increase in the ball's potential energy is 70.56 J.

Answer:

v = 2 m/s

Explanation:

Here, we will use the law of conservation of momentum to solve this problem:

[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]

where,

m₁ = mass of grandma = 80 kg

m₂ = mass of James = 40 kg

u₁ = initial speed of grandma = 3 m/s

u₂ = initial speed of James = 0 m/s

v₁ = v₂ = v = final speed of grandm and James = ?

Therefore,

[tex](80\ kg)(3\ m/s)+(40\ kg)(0\ m/s)=(80\ kg)(v)+(40\ kg)(v)\\\\(120\ kg)v = 240\ Ns\\\\v = \frac{240\ N.s}{120\ kg}\\[/tex]

v = 2 m/s

Answer:

1.66 m/s

Explanation:

Work or kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex]

[tex]4864=\frac{1}{2} (3540)v^{2}[/tex]

v = 1.66 m/s

Answer:

Explanation:

According to the statement, three confused sleigh dogs are trying to pull a sled across the Alaskan snow.  

Forces in same direction gets added , so 35N + 42N=77N  and the Net Force is 77N -53N as it is acting in opposite direction.

Net force is 25N in east to the maximum without any hassle.

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thankyou : )

The bullet stops moving on hitting on a surface. Hence, the impulse here is equal to the momentum. Therefore, the mass of the bullet is 0.1 Kg.

Impulse in physics is the change in momentum. It is the product of the force and change in time.

hence, impulse = f. dt

When the bullet is travelling with  a velocity of 500 m/s it has a momentum. When it brought to rest, momentum become zero. Thus, the momentum is equal to the impulse here.

Therefore, f.dt = m. v

f.dt = 50 N s

v = 500 m/s

m = 50 N s/500 m/s = 0.1 Kg

Therefore, the mass of the bullet is 0.1 Kg.

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Answer:

a. 6000J or 6KJ

b. Force =600

Answer:

did any of this help

Explanation:

y = (-2/3)x - 1

y-(-5)= -2/3(x-6)

y-y1=m(x-x1)

2x-3y=11

Answer:

Radioactive materials give off a form of energy called ionizing radiation.

The speed at which a clock would have to be moving in order to run at half the rate of a clock at rest depends on the theory of relativity that you are using.

In special relativity, time dilation is the phenomenon where time appears to pass differently for objects in motion relative to an observer at rest.

According to the theory, time appears to slow down for an object as it approaches the speed of light. The rate at which time appears to pass for an object is given by the equation:

T' = T / [tex]\sqrt{(1 - (v^2 / c^2))}[/tex]

Where T is the time as measured by an observer at rest, T' is the time as measured by an observer moving relative to the object, v is the velocity of the object, and c is the speed of light.

In addition, this is a theoretical scenario, practically in order to measure time dilation in a laboratory, it is required a very high precision of measurements, that are currently not possible.

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Answer:

Chemical energy to sound energy to heat energy

Answer:

Explanation:

Given that a centripetal force is a form of force that gives rise or causes a body to move in a curved path.

Hence;

1. When a car is being driven around a track, it is the FORCE OF FRICTION that is acting upon the turned wheels of the vehicle, which transforms into the centripetal force required for circular motion.

2. When a ball being is swung on the end of a string, TENSION FORCE acts upon the ball, which transforms the centripetal force required for circular motion.

3. When the moon is orbiting the earth, it is the FORCE OF GRAVITY acting upon the moon, which transforms the centripetal force required for circular motion.

4. A rotating wheel on the other hand has NO centripetal force because centripetal force is pull towards the center of a motion. However the speed of the object is tangent to the circle, while the direction of the force is also perpendicular to the direction of the rotating wheel.

The force that provides the centripetal force in each of the given situations are;

A) Friction Force

A) Friction ForceB) Tension Force

A) Friction ForceB) Tension ForceC) Force of gravity

A) Friction ForceB) Tension ForceC) Force of gravity D) No centripetal force

When an object is in circular motion, the force that keeps it moving round the circle while centrifugal force is the one that tries to pull the object away from the center.

A) When a car is driving around a track, there is a frictional force between the tires of the car and the track that acts on the vehicle to keep it in that circular motion. This frictional force is the centripetal force required to keep the vehicle in circular motion.

B) When a ball is swing on the end of a string, there is an upward force called tension force that acts on the ball to keep it swinging in circular motion. Thus, the centripetal force here is provided by the tension force.

C) When the moon is orbiting the earth, there is a force of gravity exerted by the earth on the moon that keeps the moon in a circular motion about Earth instead of moving in a straight line.

D) For a rotating wheel, the centripetal force does not do any work. The reason for that is because the centripetal force points toward the center of the circle, and as a result it means that the velocity of the rotating wheel is tangent to the circle.

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Answer:

Is at the pivot of the wheel

Answer:

a)   v = [tex]\sqrt{G \frac{M_e}{(R+R_e)} }[/tex],   b)  Em = - ½ m [tex]G \frac{M_e}{(R+R_e)}[/tex], c)  T = 2π [tex]\sqrt{\frac{ (R+R_e)^3 }{G M_e } }[/tex]

Explanation:

a) For this exercise we must use Newton's second law with the gravitational force

          F = ma

          [tex]G \frac{m M_e}{(R+R_e)^2 }[/tex] = m a

the acceleration of the satellite is centripetal

          a = [tex]\frac{v^2}{(R+R_e)}[/tex]

we substitute

            [tex]G \frac{m M_e}{(R+R_e)^2 } = m \frac{v^2}{ (R+R_e)}[/tex]

          [tex]G \frac{M_e}{(R+R_e)}[/tex]  = v²

          v = [tex]\sqrt{G \frac{M_e}{(R+R_e)} }[/tex]

the distance is from the center of the earth

b) mechanical energy is the sum of kinetic energy plus potential energy

         Em = K + U

         Em = ½ m v² - G m M / (R + R_e)

we substitute the expression for the velocity

         Em = ½ m  [tex]G \frac{M_e}{(R+R_e)}[/tex]  - [tex]G \frac{M_e}{(R+R_e)}[/tex]  

         Em = - ½ m [tex]G \frac{M_e}{(R+R_e)}[/tex]  

c) as the orbit is circulating, the velocity modulus is constant

         v = d / t

in a complete orbit the distance traveled of the circle is

        d = 2π (R + R_e)

where time is called period

         v = 2π (R + R_e)

         T = 2π (R + R_e) / v

we substitute the speed value

        T = 2π (R + R_e) [tex]\sqrt{\frac{(R+R_e) }{G M_e } }[/tex]

        T = 2π [tex]\sqrt{\frac{ (R+R_e)^3 }{G M_e } }[/tex]

(a) An expression for the speed of the satellite in its orbit.

[tex]V=\sqrt{G\dfrac{M_e}{R+R_e}[/tex]

(b) An expression for the total mechanical energy of the satellite-Earth system in its orbit.

[tex]E_m =\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}[/tex]

(c) An expression for the period of the satellite’s orbit.

[tex]T=2\pi\sqrt\dfrac{(R+R_e)^3}{GM_e}[/tex]

A satellite is a moon, planet or machine that orbits a planet or star. For example, Earth is a satellite because it orbits the sun

a) For this exercise we must use Newton's second law with the gravitational force

F = ma

[tex]ma =G\sqrt{\dfrac{mM_e}{(R+R_e)}[/tex]

the acceleration of the satellite is centripetal

[tex]a=\dfrac{v^2}{R+R_e}[/tex]

we substitute

[tex]G\dfrac{mM_e}{(R+R_e)}=m\dfrac{v^2}{(R+R_e)}[/tex]

[tex]G\dfrac{M_e}{(R+R_e)}=v^2[/tex]

[tex]v=\sqrt{G\dfrac{M_e}{(R+R_e)}[/tex]

b) mechanical energy is the sum of kinetic energy plus potential energy

        Em = K + U

[tex]Em =\dfrac{1}{2}m v^2 - \dfrac{G m M} {(R + R_e)}[/tex]

we substitute the expression for the velocity

[tex]E_m=\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}-G\dfrac{M_e}{(R+R_e)}[/tex]

[tex]E_m=-\dfrac{1}{2}G\dfrac{M_e}{(R+R_e)}[/tex]

c) as the orbit is circulating, the velocity modulus is constant

[tex]v=\dfrac{d}{t}[/tex]  

in a complete orbit the distance traveled of the circle is

[tex]d = 2\pi (R + R_e)[/tex]

where time is called period

[tex]v = 2\pi (R + R_e)[/tex]

[tex]T = \dfrac{2\pi (R + R_e)} { v}[/tex]

we substitute the speed value

[tex]T = 2\pi (R + R_e) . \sqrt{\dfrac{(R+R_e)}{GM_e}[/tex]

[tex]T=2\pi\sqrt{\dfrac{(R+R_e)}{GM_e}[/tex]

(a) An expression for the speed of the satellite in its orbit.

[tex]V=\sqrt{G\dfrac{M_e}{R+R_e}[/tex]

(b) An expression for the total mechanical energy of the satellite-Earth system in its orbit.

[tex]E_m =\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}[/tex]

(c) An expression for the period of the satellite’s orbit.

[tex]T=2\pi\sqrt\dfrac{(R+R_e)^3}{GM_e}[/tex]

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