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:

distance = 112 miles

Explanation:

its 12 miles every 0.6 in a hour

Answer:

An Oven

Explanation:

The heat is higher, so it moves faster. Shile in a freezer the particles are extremely slow!

Answer:

Explanation:

(A)

Force is said to be the rate of change of momentum.

At the instant, the parallel component of the change of momentum is zero. SInce centripetal force always acts towards the center, hence the direction upwards ( along the positive y-axis).

[tex]\dfrac{dp}{dt} = F(+\hat y)[/tex]

[tex]= \dfrac{mv^2}{r}(+\hat y) \ \\ \\ = \dfrac{28 \ kg \times (7 \ m/s)^2}{3.3 \ m }( + \hat y) \\ \\ = \mathbf{416 \ N ( + \hat y)}[/tex]

(B)

The net force acting on the child is:

[tex]F_{net} = \dfrac{mv^2}{r}( + \hat y) \\ \\[/tex]

[tex]= \dfrac{28 \ kg \times (7 \ m/s)^2}{3.3 \ m }( + \hat y) \\ \\ = \mathbf{416 \ N ( + \hat y)}[/tex]

Answer:

a = 72 m/s^2

Explanation:

Since the object is moving in circle, the acceleration points TOWARDS the centre of the circle (it is centripetal acceleration)

a = v^2 / r

a = 60^2 / 50

a = 72 m/s^2

Answer:

OMG IM ON THE SAME QUESTION

Explanation:

Answer:  I'm not sure, but I think it would be a total lunar eclipse

When the moon passes through the penumbra of Earth’s shadow it is referred to as partial lunar eclipse. The correct option is C.

A partial lunar eclipse occurs when the moon is not completely immersed in the umbra of the earth's shadow.

During a partial solar eclipse, the Moon, Sun, and Earth do not align perfectly straight, and the Moon casts only the penumbra of its shadow on Earth. From our vantage point, it appears that the Moon has eaten the Sun.

A shadow's penumbra is the lighter outer edge. Partial solar eclipses are caused by the Moon's penumbra, while penumbral lunar eclipses are caused by the Earth's penumbra. The penumbra is a type of lighter shadow.

The Penumbra is a half-shadow region that occurs when an object only partially covers a light source.

Thus, the correct option is C.

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Answer:

electric force

Explanation:

its a contact and noncontact force

Answer: 0.8 g/cm

Explanation:

p= m/V

= 4 kg/ 5 liter

= 0.8

Answer:

.8

Explanation:

give them brainliest

Answer:

Regular reflection

Explanation:

Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

i hope this helps a bit.

According to the context, a sharp image is formed when light reflects from a regular reflection.

It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.

This reflection of light happens when the angles that the two rays determine with the surface are equal.

Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.

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Answer:

Is that your answer

Answer: 10.98 N

Explanation:

Given

mass of box is [tex]m=0.5\ slugs\approx 7.3\ kg[/tex]

The coefficient of kinetic friction is [tex]\mu= 0.4[/tex]

It is given that on the application of force box started moving with constant speed i.e. there is no net external force

[tex]\Rightarrow F+mg\sin 30^{\circ}=\mu mg\cos 30^{\circ}\\\Rightarrow F=mg(\mu \cos 30^{\circ}-\sin 30^{\circ})\\\Rightarrow F=7.3\times 9.8(0.866\times0 .4-0.5)=-10.98\ N\\[/tex]

The negative sign indicates direction of force is opposite i.e. it must be applied upwards

Explanation:

रात में घूमने वाला arthaarat निशाचर

Answer:

-11.600N

Explanation:

lol

Answer:

5.0/s

Explanation:

Answer:

b and a it is this that abewsr

Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s[/tex]

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

Answer:

a) the plastic tube need to be 24.7 m long

b) the kilometer of copper wire required is 15.7

c) the resistance of this solenoid is 5538.8 2 ohms

Explanation:

Given the data in the question;

we determine the length of the plastic tube. assuming the solenoid is long.

the self inductance of a long solenoid is;

L = μ₀n²πr²l

μ₀ = 4π × 10⁻⁷ T-m/A

where

n = number of turns per unit length

r = radius of the solenoid = 8cm (as the diameter of the plastic hollow tube is 16 cm)

l = length of the solenoid or the length of the plastic tube

we find n = number of turns per unit length

given that, the copper wire to be wound around the solenoid is 0.79 mm in diameter

number of turns per meter = n = 1 / ( 0.79 × 10⁻³ m ) = 1265.8 turns/meter

So from our previous formula, we find l

L = μ₀n²πr²l

we substitute

1.0 H = (4π × 10⁻⁷ T-m/A)( 1265.8 )²(3.14)(0.08)² ( l)

1 = 0.04048 × l

l = 1 / 0.04048

l = 24.7 m

Therefore, the plastic tube need to be 24.7 m long

b)  

n = number of turns per unit length = 1265.8 turns/metre

so, the length of the plastic tube over which the copper wire is to be wound,

number of turns of copper wire required = n × l

= 1265.8 turns/meter × 24.7 m

= 31,265.26 turns

Now each turn of the copper wire is to be wound across the 18cm diameter of the plastic tube.

so for each turn length of copper wire required = 2π × r

= 2π × 0.08 m

= 0.5026548 m

So copper wire required for 31,265.26 turns will be;

⇒ 31,265.26 × 0.5026548 = 15715.63m = 15.7 km

Therefore, the kilometer of copper wire required is 15.7

c)

p = resistivity of copper = 1.68 × 10⁻⁸ ohm-m

Resistance = pl/a

where l is length of copper wire, a is cross sectional area;

diameter of copper wire is 0.79-mm

radius of copper wire is 0.79/2 = 0.395 mm = 0.000395 m

area of cross section of copper wire a = πr² = π( 0.00395)² = 4.9 × 10⁻⁷ m²

Resistance = pl/a

we substitute

Resistance = [(1.68 × 10⁻⁸ ohm-m)( 15715.63m )] / [ 4.9 × 10⁻⁷ m² ]

Resistance =  5538.8 2 ohms

Therefore,  the resistance of this solenoid is 5538.8 2 ohms

Explanation:

When hot water is poured on the can in a bucket of cold water, the can crushes off means it gets unshaped

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

   = [tex]130\times 6.894[/tex]

   = [tex]896.22 \ Kpa[/tex]

or,

   = [tex]896.22\times 10^3 \ Pa[/tex]

Z₂ = 10ft

    = 3.05 m

[tex]\delta[/tex] = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒  [tex]\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2[/tex]

On substituting the values, we get

⇒  [tex]\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2[/tex]

⇒  [tex]\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05[/tex]

⇒  [tex]P_2=866330 \ P_a[/tex]

i.e.,

⇒       [tex]=866330\times 0.000145[/tex]

⇒       [tex]=126 \ Psi[/tex]

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:

they are right it is a new moon

Explanation:

took the test

Answer:

320 J

Explanation:

From the question,

KE = 1/2mv².................. Equation 1

Where KE = Kinetic Energy, m = mass of the rock, v = velocity of the rock

Given: m = 40 kg, v = 4m/s

Substitute these values into equation 1

KE = 1/2(40)(4²)

KE = 20×16

KE = 320 J

Hence the kinetic energy of the rock is 320 J

Answer:

The correct answer is A) It is upright and 1.5m behind the mirror

Explanation:

Your reflection must be upright, meaning vertical/erect, and the distance will be the exact same. Also, the reflected ray appears as if it had traveled from an object located behind the mirror.

Answer:

If the angular acceleration is a constant A, then this constant times the time will be equal to the angular velocity.

We know that after accelerating for 10 seconds, the angular velocity is 0.4 rad/s

Then:

A*10s = 0.4 rad/s

If we solve this for A we get:

A = (0.4 rad/s)/10s = 0.04 rad/s^2

This is the angular acceleartion.

Now we assume that the angular speed remains constant at 0.4 rad/s, and we know that the radius of the carousel is 20m, we want to find the linear velocity of the children.

We know that for an angular velocity W and for a radius R, the linear velocity is:

V = R*W

Then in this case the linear velocity is:

V = 20m*(0.4 s^(-1)) = 8 m/s

Answer:

The respiratory system provides oxygen for cells, while the circulatory system transports oxygen to cells.

Explanation:

so the answer is D

Answer:

20 J/s

Explanation:

Given data

Kinetic energy=100 J

Time= 5 seconds

Hence the rate at which the kinetic friction work is

=100/5

=20 J/s

Therefore the answer is 20 J/s

Kinetic energy decreases if either the mass or the velocity of the object decreases or if both decrease. and  Kinetic energy increases if either the mass or the velocity of the object increases or if both increase.

Explanation:

These statements explain the relationship between mass, velocity, and kinetic energy .i.e KE = ½mv², (3) Kinetic energy decreases if either the mass or the velocity of the object decreases or if b decrease. (4)Kinetic energy increases if either the mass or the velocity of the object decreases or if both decrease.

When a body has a mass of m kg and travels a velocity of v m/s then the kinetic energy is given by,

KE=1/2mv²

In the above relation, it is clearly shown that,

Kinetic energy is directly proportional to the mass of the body,

i.e

KE∝m

and also Kinetic energy directly varies with the square of the velocity of the body.

I.e

KE∝v²

Therefore, From the given expression we can conclude that when mass and velocity increased either one of them or both, the kinetic energy also gets increased.

And when mass and velocity decreased either one of them or both, the kinetic energy also gets decreased. Hence options (3) and (4) are correct.

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Answer:

the oscillations of the electrons must be in the 10⁸ Hz = 100 MHz range

Explanation:

The speed of a wave of radio, television, light, heat, all are manifestations of electromagnetic waves that are oscillations of electric and magnetic fields that support each other, the speed of all these waves is the same and the vacuum is equal to c = 3 108 m / s

All waves have a relationship between the speed of the wave, its frequency and wavelength

          c = λ f

          f = c /λ

for this case lam = 1 m

          f = 3 10⁸/1

          f = 3 10⁸ Hz

the oscillations of the electrons must be in the MHz range

It should be clarified that the speed of light in air is a little lower

          n = c / v

          v = c / n

the refractive index of vacuum is n = 1 and the refractive index of air is n = 1.000002

Answer:

hydrogen, helium, lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, neon, sodium, magnesium, aluminum, silicon, phosphorus, sulfur, chlorine, argon, potassium, calcium, scandium, titanium, vanadium, chromium, maganese, iron, cobalt, nickel, copper, zinnc, gallium, germanium, aresnic, selenium, bromine, krypton, rubidium, strongtium, yttrium, zirconium, niobium, molybdenum, technetium, ruthenium, rhodium, palladium, silver, cadmium, indium, tin, atimony, tellurium, iodine, xenon, cesium, barium, lanthanum, cerium, praseodyumium, neodymim, promethium, samarium, europium, gadolinium, terbium, dysprosium, holmium, erbium, thulium, ytterbium, luteium, hafnium, tantalum, tungsten, rhenium, osmium, irdium, platinum, gold, mercury, thallium, lead bismuth, polnium, astatine, radon, rancium, radium, actinum, thorium, protactinium

Explanation: