If we always drop the balls from 1-m
height in each trial, what type of variable
is this in this experiment?
Constant Variable
Dependent Variable
Independent Variable

Answer:

bb kettle

Explanation:

it transfres electricsl to kinetic

Answer:

OMG IM ON THE SAME QUESTION

Explanation:

Answer:

answer is 3

Explanation:

by using s= [(v+ u)/2] x t

28= (v+ 0)/2 x 11

v= 5.09 ms^-1

Answer:

1) 1000Nm

2)  95,625Nm

3) 1.05%

Explanation:

Mechanical Advantage is the ratio of the load  to the effort applied to an object.

MA = Load/Effort

1) Workdone on the load = Force(Load) * distance covered by the load

Workdone on the load = 500N * 2m

Workdone on the load = 1000Nm

2)  work done by the effort = Effort * distance moves d by effort

work done by the effort = 2125 * 45

work done by the effort = 95,625Nm

3) Efficiency = Workdone on the load/ work done by the effort * 100

Efficiency = 1000/95625 * 100

Efficiency = 1.05%

Hence the efficiency of the system is 1.05%

The answer is Motion

Answer:

False

Explanation:

I learned it the hard way trust me T^T

Answer:

materials that allow electricity to pass through them are called conductors, some examples of conductors are many metals, such as copper, iron and steel.

Answer:

1115560000 J

Explanation:

1/2 * 80,000 * 167^2 m/s = 1115560000 J

Answer:

a) the torque required is  10.53 N-m

b) The magnitude force applied tangentially is 12.33 N

Explanation:

Given the data in the question;

mass m = 1.765 kg

radius r = 0.854 m

first we calculate the moment of inertia;

[tex]I[/tex] = [tex]\frac{2}{5}[/tex]mr²

we substitute

[tex]I[/tex] = [tex]\frac{2}{5}[/tex] × 1.765 × (0.854)²

[tex]I[/tex] = 0.514897 kg.m²

a)

Find the torque required to bring the sphere from rest to an angular velocity of 317 rad/s, clockwise, in 15.5 s

ω[tex]_{initial[/tex] = 0

ω[tex]_{final[/tex] = 317 rad/s

t = 15.5 s

we know that; ω[tex]_{final[/tex] = ω[tex]_{initial[/tex] + ∝t

so we substitute

317 = 0 + ∝(15.5)

∝ = 317 / 15.5

∝ = 20.4514 rad/s²

so

ζ = [tex]I[/tex] × ∝

we substitute

ζ = 0.514897 × 20.4514

ζ = 10.53 N-m

Therefore, the torque required is  10.53 N-m

b)

What magnitude force applied tangentially at the equator would provide the needed torque.

ζ = F × r

we substitute

10.53  = F × 0.854

F = 10.53 / 0.854

F = 12.33 N

Therefore, magnitude force applied tangentially is 12.33 N

Answer:

Is where two or more inductors are “linked” so that voltage is induced in one coil proportional to the rate-of-change of current in another

Answer:

151.12 million miles i actually just did this question on a test

Explanation:

Mars orbits—or completes one revolution—around the Sun every 686.98 Earth days, or once every 1.88 Earth years.

The term "revolution" describes how an item moves in its orbit around another object. For instance, the 24-hour day is created as the Earth rotates on its own axis.

The 365-day year is made possible by the Sun's rotation of the Earth. A planet spins around a satellite.

While orbiting the Sun, Mars travels at an average speed of 53,979 miles per hour, which equates to 86,871 km per hour.

Thus, Mars orbits—or completes one revolution—around the Sun every 686.98 Earth days, or once every 1.88 Earth years.

To learn more about One revolution, refer to the below link:

https://brainly.com/question/26289977

# SPJ2

Answer:

The first harmonics are the fundamental, the first overtone, and the second overtone,

For a string fixed at both ends you have:

N-A-N            F/2

N-A-N-A-N     F

N-A-N-A-N-A-N    3 F/2        each harmonic increases by F/2

Where N are the anti-nodes and A the antinodes

N-A-N-A-N    represents 1 wavelength for string fixed at both ends

So 1 wavelength is 64 cm

280 m/s = F * .64 m       and F = 437.5 Hz       (second harmonic)

First harmonic = F/2 = 218.75 / sec

Third harmonic = 3 F / 2 = 656.25 /sec

Answer: 0.8 g/cm

Explanation:

p= m/V

= 4 kg/ 5 liter

= 0.8

Answer:

.8

Explanation:

give them brainliest

Answer:

[tex]3.52\times 10^{25}\ \text{J}[/tex]

Explanation:

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]

m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]

[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]

[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]

Energy required is given by

[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]

The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].

Answer:

B or 2

Explanation:

Answer:

The correct answers are: A, C, D, E

Explanation:

The magnetic field is a solenoid is given by

         B = μ₀  [tex]\frac{N}{L}[/tex]   I

where N is the number of turns, I the current and L length of the solenoid.

Using this equation let's examine the different responses to permute increasing the magnetic field

A) True. a thicker wire decreases the resistance and the current can increase the system.

B) False. If there is no voltage source there is no current in the system

C) True. the field is proportional to the number of turns

D) True. the magnetic moments of the core align with the field increasing its value

E) True. When the loops are closer together, more of them can fit per unit length

F) False. If the wire is shorter the number of turns decreases.

The correct answers are: A, C, D, E

Answer :

Last choice

Answer:

its frequency is too low

Answer:

magnitude is 1382.59 N/C

Explanation:

Given the data in the question;

The time taken is;

t = x / v

we substitute;

t = ( 2 × 10⁻²) / ( 5.69  × 10⁶ )

t = 3.5149 × 10⁻⁹ s

next, the acceleration is;

a = 2y/t² = [2( 0.150 × 10⁻²)] / [ ( 3.5149 × 10⁻⁹ )² ]

a =  2.42826 × 10¹⁴ m/s²

now, the electric field is;

E = ma / q

we know that;

mass of electron m = 9.11 × 10⁻³¹ kg,

charge of electron q = 1.60 × 10⁻¹⁹ coulomb

we substitute

E = ( 9.11 × 10⁻³¹ )(2.42826 × 10¹⁴) / 1.60 × 10⁻¹⁹

E = 2.21214 × 10⁻¹⁶  / 1.60 × 10⁻¹⁹

E = 1.3826 × 10²¹

E = 1382.59 N/C

Therefore, magnitude is 1382.59 N/C  

Answer:

15m

Explanation:

Given that a roller coaster starts from rest at A, rolls down the track to B, describes a circular loop of 12-m diameter, and travels up and down past point E. Determine the range of values of h for which the roller coaster will not leave the track at D or E. Assume no energy loss due to friction

Solution

At point A

The maximum potential energy = maximum K.E

At point A, the total energy = maximum P.E.

Down the track to point B, the P.E will be converted to maximum K.E.

Hence,

Mgh = 1/2mv^2

Also, the total energy at the roller coaster will be P.E + K.E

I.e mg2r + 1/2mv^2

Where 2r = height of the loop = diameter of the loop.

Since the energy is always conserved, hence

Mgh = mg2r + 1/2mv^2

Let also consider the centripetal acceleration to keep the object in the circle.

F = mV^2 / r = mg

Mass will cancel out

U^2 = rg

Substitute that in the last equation

Mgh = mg2r + 1/2mgr

mgh = mg ( 2r + 1/2r )

Mg will cancel out

h = 2.5r

Where r = 12/2 = 6

h = 2.5 × 6

h = 15m

Therefore, the values of h for which the roller coaster will not leave the track at D or E is 15m.

Answer:

the required frequency of waves is 2.066 Hz

Explanation:

Given the data in the question;

μ = 1.50 kg/m

T = 6000 N

Amplitude A = 0.500 m

P = 2.00 kW = 2000 W

we know that, the average power transmit through the rope can be expressed as;

p = [tex]\frac{1}{2}[/tex]vμω²A²

p = [tex]\frac{1}{2}[/tex]√(T/μ)μω²A²

so we solve for ω

ω² = 2P / √(T/μ)μA²

we substitute

ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²

ω² = 4000 / 23.71708

ω² = 168.65

(2πf)² = ω²

so

(2πf)² = 168.65

4π²f² = 168.65

f² = 168.65 / 4π²

f² = 4.27195

f = √4.27195

f = 2.066 Hz

Therefore, the required frequency of waves is 2.066 Hz

Answer:

a new chromebook for you and you will get to know the other one that

Answer:

0.318cm

Explanation:

The computation of the radius of the wheel is shown below:

As we know that

One revolution of the wheel = 2πr

Here r denotes the radius of the wheel

Now

5 revolutions of the wheel would be

= 2πr ×  5

= 10πr

So,

10πr = 10cm

Thus r = 1 ÷ π

= 0.318cm

The radius of the wheel will be r=0.318 cm

The radius is defined as the distance between the center of circle and the outer layer of the circle.

It is given in the question that

Revolutions of the wheel = 5

The length of the string =10 cm

Now the radius will be calculated as

One revolution of the wheel

= [tex]2\pi r[/tex]

Now 5 revolutions of the wheel would be

=  [tex]2\pi r\times 5=10\pi\ r[/tex]

Since the string has the length of 10 cm then

[tex]10\pi\ r=10[/tex]

[tex]r=\dfrac{10}{10\pi}[/tex]

[tex]r=0.318\ cm[/tex]

Thus the radius of the wheel will be r=0.318 cm

To know more about radius of circle follow

https://brainly.com/question/977264

Answer:

There will be no tides

Explanation: