A coin is located 35.0 mm from the center of a turntable which accelerates from rest to a speed of 7.95 rad/s in 0.270 s. What is the minimum coefficient of friction to avoid slippage of the coin

True, the electric field produced by a uniform, infinite plane of charge does not depend on the distance from the plane.

Electric field is the region of space at which electric force is felt.

E = kq/r²

where;

at infinite plane, r → ∞, and E = 0

Thus, the electric field produced by a uniform, infinite plane of charge does not depend on the distance from the plane.

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Answer:

[tex]2000\; {\rm cm^{3}}[/tex].

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let [tex]m(\text{ball})[/tex] denote the mass of this ball. Let [tex]m(\text{water})[/tex] denote the mass of water that this ball has displaced.

Let [tex]g[/tex] denote the gravitational field strength. The weight of this ball would be [tex]m(\text{ball}) \, g[/tex]. Likewise, the weight of water displaced would be [tex]m(\text{water})\, g[/tex].

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

[tex]\text{buoyancy} \ge m(\text{ball})\, g[/tex].

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

[tex]\text{buoyancy} = m(\text{water}) \, g[/tex].

Therefore:

[tex]m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g[/tex].

[tex]m(\text{water}) \ge m(\text{ball})[/tex].

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let [tex]\rho(\text{water})[/tex] denote the density of water. The volume of water that this ball should displace would be:

[tex]\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \end{aligned}[/tex].

Given that [tex]m(\text{ball}) = 2000\; {\rm g}[/tex] while [tex]\rho = 1.00\; {\rm g\cdot cm^{-3}}[/tex]:

[tex]\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}[/tex].

In other words, for this ball to stay afloat, at least [tex]2000\; {\rm cm^{3}}[/tex] of the volume of this ball should be under water. Therefore, the volume of this ball should be at least [tex]2000\; {\rm cm^{3}}\![/tex].

Answer:The velocity of the object will be 5[tex]\sqrt7[/tex]m/s or 13.23m/s

Explanation:

force exerted by the object= 30N

distance displayed by the object by the action of force=6.0m

mass of object=10kg

velocity gained by the object=?

[tex]\frac{1}{2}mv^{2}= forcexdisplacement\\\frac{1}{2}10v^{2} = 30x6\\ 5v^{2}=180\\ v^{2}= 180-5\\ v^{2} =175\\v=\sqrt{175} \\v=5\sqrt{7} or 13.23[/tex]

Answer:

75%

Explanation:

Efficiency=MA/VR*100

=3/4*100

=0.75*100

=75%

a)The ball's potential energy changes as it moves from the release point to the top of the rebound.

b)The ball's potential energy is at its highest and rests when it is in the top position.

The potential energy is due to the virtue of the position and the height. The unit for the potential energy is the joule.

The potential energy is mainly dependent upon the height of the object.

Potential energy = mgh

The kinetic energy of the body is due to the virtue of motion.

According to the Law of Conservation of Energy, energy can neither be created nor destroyed but can be transferred from one form to another.

The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

A ball dropped onto a trampoline returns to the same height after the rebound.

Hence, the ball's potential energy changes as it moves from the release point to the top of the rebound. When the ball is at the top position the potential energy is maximum and the ball is at rest.

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Answer:

Density is defined as mass per unit volume. its SI unit is: kg/m³

Explanation:

Answer:

Density is defined as mass per unit volume. its SI unit is: kg/m³

Explanation:

from answer above

Answer:

34g

(second option listed)

Explanation:

half life is the number of days it takes to decrease by half,

after 8 days you are left with half of the original portion

in a period of 32 days, we experience 4 repeats of 8 days

(it gets cut in half four times):

544 / 2 = 272  {day 8}

272 / 2 = 136  {day 16}

136 / 2 = 68 {day 24}

68 / 2 = 34 {day 32}

so, after 32 days, 34g of ooblek are left un-decayed

hope this helps!! have a lovely day :)

Answer:

B

Explanation:

The portions of Earth's fresh water are located in lakes and streams.  The options A and D are correct.

The freshwater is the one which does not contains any harmful chemical and saltiness.

The oceans contains salt water and the lakes and the streams have fresh water.

So, the option A and D are correct.

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Answer:

He is being careless.

Explanation:

he is trying to "get rid" of one problem so he can tackle another. This is food though, and you have customers to serve. You also have to be careful about food poisoning.

The peak value of a sine wave whose VAV value is 15V is 21. 21

A peak value is simply known as the highest or maximum value reached in one cycle by an alternating quantity.

The formula for peak value is given as;

peak value = Voltage * √2

Voltage = 15V

Peak value = 15 * √ 2 = 15 * 1. 4142 = 21. 21

Thus, the peak value is 21. 21

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From the calculations, the force required is 0.437 N.

The term force is defined a the product of mass and acceleration. The force is required to move the sling the required distance.

If mass = 0.019 kg and acceleration =  23 m/s2 then we know that F = ma

It then follows that the required force is;

F = 0.019 kg  *  23 m/s2= 0.437 N

Missing parts

You plan to use a slingshot to launch a ball that has a mass of 0.019 kg. You want the ball to accelerate straight toward your target at 23 m/s2. How much force do you need to apply to the ball?

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Ridge push refers to the pressure exerted by the excess height of the mid-ocean ridge, while Slab pull refers to the force exerted by the weight of the subducted slab on the plate it is attached to.

The two forces that form mid-ocean ridges are ridge-push and slab pull.

Ridge push refers to the pressure exerted by the excess height of the mid-ocean ridge.

Slab pull refers to the force exerted by the weight of the subducted slab on the plate it is attached to.

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Suppose the resistance of a copper wire is 1.05 ohms at 30°C. the temperature is mathematically given as

T=139.89

Generally, the equation for the coefficient of change is mathematically given as

[tex]\mu_k= 0.2+(0.043)3\\\\\mu_k=0.2+(0.043)(4.343)[/tex]

[tex]\mu_k=0.387[/tex]

In conclusion, the Temperature of copper

[tex]R=R_0( 1+ \alpha(T-T1))[/tex]

1.5(1+0.005(T-30))

T=139.89

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Answer:

Approximately [tex]122.625\; {\rm m}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex], the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let [tex]v_{0}[/tex] denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly [tex](-v_{0})[/tex]. The velocity of the ball would be changed from [tex]v[/tex] to [tex](-v_{0})\![/tex] (such that [tex]\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})[/tex]) within [tex]t = 10\; {\rm s}[/tex].

Also because the drag on the ball is negligible, the acceleration of the ball would be [tex]a = -g = -9.81\; {\rm m\cdot s^{-2}}[/tex]. Thus:

[tex]\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}[/tex].

Since [tex]\Delta v = (-2\, v_{0})[/tex]:

[tex]-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}[/tex].

[tex]\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].

The ball reaches maximum height when its velocity is [tex]v_{1} = 0\; {\rm m\cdot s^{-1}}[/tex]. Apply the SUVAT equation [tex]x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)[/tex] to find the displacement [tex]x[/tex] between the original position (ground level, where [tex]v_{0} = 49.05\; {\rm m\cdot s^{-1}}[/tex]) and the max-height position of the ball (where [tex]v_{1} = 0\; {\rm m\cdot s^{-1}}[/tex].)

[tex]\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

The answer to the correct number of significant figures is 6.774.

When a number(big) divided smaller number, the answer obtained greater than zero is called a quotient.

Divide 143.6 ÷ 21.2

143.6/21.2 = 1436/212

                  =6.77358

The quotient is rounded to three significant figures after decimal

143.6 ÷ 21.2 = 6.774

Thus, the answer to the correct number of significant figures is 6.774

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Answer:

Sonam will jump further than wangchuck

Answer:

first answer:He can promote healthier start in the children's life by trying to avoid what he encounter when he has no access to health care.

Answer:

Explanation:

The two examples of contact forces are frictional force and applied force. The two examples of noncontact forces are gravitational force and magnetic force.

Answer:

kinetic energy of an object is proportional to the square root of its boiling vocelity

Answer:

I think its either A or B. I’m not really sure, don’t listen to me. Read my explanation and it should help you.

Explanation:

When the two conducting spheres are brought together to touch, it is as though they become one single big conductor and the total charge of the two spheres spreads out across the whole surface of the touching spheres. When the spheres are moved apart again, each one is left with half of the total original charge…

X and Y are two identical conducting spheres separated by a distance d. X has a charge of +6µC and Y has a charge of -2µC. The electric force between them is +F (i.e. attractive). The spheres are touched together and are then returned to their original separation d. What is the new value of the force between them? Any help is greatly appreciated. I assumed that since the net charge should remain the same after, there would be no change in the value of the force, but the answer scheme gives the answer as -F/3…

Answer:

i am pretty sure its 15 m

Explanation:

10 m/s

1.5 seconds

25 angle

10m/1.5s

10x1.5=15

electrons can be knocked loose from one object and picked up by the other. The object that gains electrons becomes negatively charged, while the object that loses electrons becomes positively charged.

b gained electrons

Answer:

1). Brake pedal or lever

2). A pushrod also called an actuating rod

The air space would be half the size it was at the surface.

It should be noted that airspace means the portion of the atmosphere controlled by a country that's above its territory.

Based on the information given, the The air space would be half the size it was at the surface.

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The mobility and hall coefficient willl be 57.7 ×10⁻⁴ and 29.41 ×10⁻⁶ v.

Electric current is defined as the ratio of the charge flow in a given time interval. It is denoted by I. It has the unit is ampere.

Given data;

Electric current,I= 25 A

Thickness  = 0.1 mm

Length = 3 mm

Flux  = 1.4 w/m²

The charge density is found as;

[tex]\rm \rho =- \frac{1}{R_H} \\\\ \rho = 0.119 \times 10^{11} C/m^3[/tex]

The value of the hall voltage is;

[tex]\rm V_H = BVd \\\\ V_H = B \frac{I}{\rho dt} d \\\\ V_H = \frac{BI}{\rho t } \\\\ V_H = \frac{0.14 \times 25 }{0.119 \times 10^{11} \times 0.1 \times 10^{-3}} \\\\ V_H = \frac{3.5 }{0.119 \times 10^7} \\\\ V_H = 29.41 \times 10^{-6} \ v[/tex]

The mobility of electrons is;

[tex]\rm \mu_e = \frac{\rho}{n_e}\\\\ \mu_e = \frac{6.848 \times 10^7}{13.5 \times 10^{-30}\times 1.607 \times 10^{-19}} \\\\ \mu_e = 57.7 \times 10^{-4}[/tex]

Hence the mobility and hall coefficient willl be 57.7 ×10⁻⁴ and 29.41 ×10⁻⁶ v.

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If you float a piece of wood in a container half-full of water that rests on a weighing scale, the weight reading on the scale will increase.

Buoyancy is the force through your fluids or total or submerged bodies. Also known as Archimedes' theorem. The fluid of the direction on the body produces a resultant force but with the direction of the weight, with the opposite direction, upwards.

All material submerged in a liquid under the action of buoyancy, for this reason we have that the value on the scale will be increased with a system of a wood floating in water.

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The object which follows Newton's first law of motion around me is; my scientific calculator.

The concept which describes objects as forever stationary except acted upon by external forces can be attributed to.the Newton's first law of motion. The postulation of the law is that; An object remains at rest if at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force. Hence, except acted upon by an external force(myself), my calculator will remain at rest.

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The objects around us that obeys newton's first law of motion are

Newton's first law of motion states that every object will continue to be in its state of rest on remain in uniform motion except it is acted upon by an external force.

Newton's first law of motion is also called the law of inertia

Examples of  newton's first law of motion are

etc.

Hence, The objects around us that obeys newton's first law of motion are

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The forces of attraction between water molecules and the glass walls and within the molecules of water themselves are what enable the water to rise in a thin tube immersed in water.

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

Surface or interfacial forces lead to capillarity. The forces of attraction between the water molecules and the glass walls and among the water molecules themselves are what causes the water in a thin tube submerged in water to rise.

Hence, the water rises up a thin capillary tube can be explained by Newton's third law.

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