Which of the following items is an example of a symbol Bernice bobs her hair

Answers

Answer 1

what? Is there a picture you can show us with this question


Related Questions

According to Newton’s law of universal gravitation, which statements are true?

Answers

1,3,5 it should be right because i have took that thing before

Find the ratio of speeds of a proton and an alpha particle accelerated through the same voltage, assuming nonrelativistic final speeds. Take the mass of the alpha particle to be 6.64 ✕ 10−27 kg.

Answers

Answer:

The required ratio is 1.99.

Explanation:

We need to find the atio of speeds of a proton and an alpha particle accelerated through the same voltage.

We know that,

[tex]eV=\dfrac{1}{2}mv^2[/tex]

The LHS for both proton and an alpha particle is the same.

So,

[tex]\dfrac{v_p}{v_a}=\sqrt{\dfrac{m_a}{m_p}} \\\\\dfrac{v_p}{v_a}=\sqrt{\dfrac{6.64\times 10^{-27}}{1.67\times 10^{-27}}} \\\\=1.99[/tex]

So, the ratio of the speeds of a proton and an alpha particle is equal to 1.99.

A positively charged plastic ruler is brought close to a piece paper resting on the desk. The piece of paper was initially neutral. When the ruler was brought closer, the paper is attracted to the ruler. The surface of the paper became charged through:_________

Answers

Answer: static electricity

Explanation:

When the plastic ruler is rubbed, friction opposes the motion and causes the transfer of electron from one surface to another such that plastic becomes negatively charged. When ruler is brought nearer to the paper, it induces the   positive charge in the piece of paper.

Electromagnetic radiation from a 8.25 mW laser is concentrated on a 1.23 mm2 area. Suppose a 1.12 nC static charge is in the beam, and moves at 314 m/s. What is the maximum magnetic force it can feel

Answers

Answer:

The maximum magnetic force is 2.637 x 10⁻¹² N

Explanation:

Given;

Power, P = 8.25 m W = 8.25 x 10⁻³ W

charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C

speed of the charge, v = 314 m/s

area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²

The intensity of the radiation is calculated as;

[tex]I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2[/tex]

The maximum magnetic field is calculated using the following intensity formula;

[tex]I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T[/tex]

The maximum magnetic force is calculated as;

F₀ = qvB₀

F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)

F₀ = 2.637 x 10⁻¹² N

PLEASE HELP ME WITH THIS ONE QUESTION
The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)

A) 2.81 eV

B) 3.89 eV

C) 2.10 eV

D) 2.78 eV

Answers

The color orange has a wavelength of 590 nm. The energy of an orange photon is approximately 0.337 eV.

The correct answer is option E.

To calculate the energy of a photon, we can use the equation:

E = (hc) / λ

where E is the energy of the photon, h is the Planck's constant (6.626 x [tex]10^-^3^4[/tex]J·s or 6.626 x[tex]10^-^1^9^[/tex] eV·s), c is the speed of light (3.00 x [tex]10^8[/tex] m/s), and λ is the wavelength of the light.

Given that the wavelength of orange light is 590 nm (or 590 x [tex]10^-^9[/tex]m), we can substitute the values into the equation:

E = [(6.626 x[tex]10^-^1^9^[/tex] eV·s) x (3.00 x [tex]10^8[/tex] m/s)] / (590 x[tex]10^-^9[/tex]m)

E = (1.9878 x [tex]10^-^1^0[/tex]eV·m) / (590 x [tex]10^-^9[/tex] m)

E = 3.3695 x [tex]10^-^1[/tex] eV

For more such information on: wavelength

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The question probable may be:

The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x [tex]10^-^1^9^[/tex], 1 eV = 1.6 x[tex]10^-^1^9^[/tex]J)

A) 2.81 eV

B) 3.89 eV

C) 2.10 eV

D) 2.78 eV

E)  0.337 eV

A 35 kg child slides down a playground slide at a constant speed. The slide has a height of 3.8 m and is 8.0 m long. Find the magnitude of the kinetic friction force acting on the child.

Answers

Answer:

The magnitude of the kinetic frictional force acting on the child is 162.93 N

Explanation:

Given;

mass of the child, m = 35 kg

height of the slide, h = 3.8 m

length of the slide, d = 8.0 m

The change in thermal energy associated with the kinetic frictional force is calculated as follows;

[tex]\Delta E_{th} + \Delta K.E + \Delta U = 0\\\\\Delta E_{th} + (\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2) + (mgh_f - mgh_i) =0\\\\since \ the \ speed \ is \ constant, \ v_f = v_i \ and \ \Delta K.E = 0\\\\Also, \ final \ height \ , h _f= 0\\\\\Delta E_{th} - mgh_i = 0\\\\\Delta E_{th} = mgh_i\\\\\Delta E_{th} = 35 \times9.8 \times 3.8\\\\\Delta E_{th} = 1303.4 \ J[/tex]

The magnitude of the kinetic frictional force that produced this thermal energy is calculated from the work done by frictional force;

[tex]\Delta E_{th} = F \times d\\\\F = \frac{\Delta E_{th} }{d} \\\\F = \frac{1303.4}{8} \\\\F = 162.93 \ N[/tex]

Therefore, the magnitude of the kinetic frictional force acting on the child is 162.93 N

PLEASE HELP ME WITH THIS ONE QUESTION
Given the atomic mass of Boron-9 is 9.0133288 u, what is the nuclear binding energy of Boron-9? (Mproton = 1.0078251, Mneutron = 1.0086649, c^2 = 931.5 eV/u)

A) 59 eV

B) 58 eV

C) 57 eV

D) 56 eV

Answers

Answer:

a. 59 ev. helpful answer

what is the difference between VELOCITY and SPEED?​

Answers

Answer:

Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement. Put another way, speed is a scalar value, while velocity is a vector. ... In its simplest form, average velocity is calculated by dividing change in position (Δr) by change in time (Δt).

Explanation:

. A car increases velocity from 20 m/s to 60 m/s in a time of 10 seconds. What was the acceleration of the car?

Answers

Answer:

0.3333

Explanation:

Acceleration = change in velocity/time

a = 20 m/s / 60 m/s

a = 0.3333 m/s^2

The large scale structure of the universe has been carefully mapped using redshift surveys of a very large number of galaxies. Answer the following: Suppose that a large telescope with modern equipment can measure the redshift to a galaxy in just 10 minutes. And suppose we want to spend no more than a year mapping the distribution of galaxies. How many redshifts can be surveyed

Answers

Answer:

26280

Explanation:

In current time, good telescope can measure redshift to a galaxy in 10 minutes.  

Thus, in one year that has on an average 365 days, the total time taken to  measure redshifts is = ( 365 *12 *60) minute  

= 262800 minutes .

Hence, the number of redshifts observed in a year = (262800/10) = 26280

Using pascals principle, F1/A1=F2/A2, solve this like a proportion.
A force of 50 N is applied to an area of 200 sq feet, how much force will be applied of the area to be covered is 50 sq feet?

Answers

Answer:

NO CLUE

Explanation:

GOOD LUCK THOUGH

26.
Which one of the following is not a vector quantity?
(2)
A) acceleration
C) displacement
E) instantaneous velocity
B) average speed
D) average velocity​

Answers

Answer:

Answer: Speed is not a vector quantity. It has only magnitude and no direction and hence it is a scalar quantity.

A spring with a 10-kg mass and a damping constant 15 can be held stretched 2 meters beyond its natural length by a force of 6 newtons. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity.

Required:
Find the position of the mass at any time t.

Answers

Answer:

Explanation:

Given that:

mass = 10kg

damping constant C = 15 kg/s

length = 2 m

Force F = 6N

Using the Hooke's law:

F = kx

6 = 15x

k = 6 N /2 m

spring constant k = 3 N/m

For the critical damping

C² - 4k*m= 0

m = C²/4k

m = (15)²/4(3) kg

m = 225/12 kg

m = 18.75 kg

A 56 kg pole vaulter falls from rest from a height of 5.1 m onto a foam rubber pad. The pole vaulter comes to rest 0.29 s after landing on the pad.

Required:
a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision

Answers

a. The athlete's velocity just before reaching the pad is [tex]35.21m/s[/tex]

b. The constant force exerted on the pole vaulter is 6799.52 N

a. We use Newton's equation of motion,

                    [tex]v=u+at\\\\S=ut+\frac{1}{2}at^{2}[/tex]

Where u is initial velocity, v is final velocity, a is acceleration , t is time and S represent distance.

Given that,  s = 5.1 m , t = 0.29s, u = 0

Substitute in above equation.

            [tex]5.1=\frac{1}{2}*a*(0.29)^{2} \\\\a=\frac{5.1*2}{0.084}=121.42m/s^{2}[/tex]

the athlete's velocity, [tex]v=0+121.42*(0.29)=35.21m/s[/tex]

b. The constant force exerted on the pole vaulter due to the collision is given as,             [tex]Force=mass*acceleration[/tex]

             [tex]Force=56*121.42=6799.52N[/tex]

Learn more:

https://brainly.com/question/13532462

The correct formula for finding the relative velocity of an object is:

WILL MARK BRAINLIEST TO THE CORRECT ANSWER!!

Answers

Answer:

[tex]V_{a/c} = V_{a/b} + V_{b/c}[/tex]

Explanation:

The relative velocity of an object is the velocity of the object relative to the observer or frame of reference.

The velocity of particle "A" with respect to particle "B" is written as [tex]V_{A/B} = V_{A} - V_{B}[/tex]

From the given options, the second option is the correct answer.

[tex]V_{a/c} = V_{a/b} + V_{b/c}\\\\Re-arrange \ the \ above \ equation;\\\\V_{a/c} - V_{b/c}= V_{a/b}\\\\or\\\\V_{a/b}= V_{a/c} - V_{b/c}[/tex]

A gymnast of mass 70.0 kgkg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81m/s29.81m/s2 for the acceleration of gravity.
PART A Calculate the tension T in the rope if the gymnast climbs the rope at a constant rate.
PART B Calculate the tension TTT in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.00 m/s2
PART C Calculate the tension TTT in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.00 m/s2m/s2 .

Answers

Answer:

43994

Explanation:

Hope this helps!

B. Complete the lists:
Things that I must do for my family
Things I must never do to my family
1.
2.
2.
3.
3.
4.
5.
5.​

Answers

Answer:

Things you should do for your family

help your parentstreat them kindlylisten and obey themappreciate them for anything they do for you talk softly

things you shouldn't

backanswering them Disobey And anything that's harsh or make it parents sad

Which statement describes an action-reaction pair?
O A. You push on a car, and the car pushes back on you.
B. A book pushes down on a table, and the table pushes down on the
Earth.
C. The Moon pulls on Earth, and Earth pulls on the Sun.
D. You push down on your shoe, and Earth's gravity pulls down on the
shoe.

Answers

Answer:

A

Explanation:

a pex

A mass of 4 kg is traveling over a quarter circular ramp with a radius of 10 meters. At the bottom of the incline the mass is moving at 21.3 m/s and at the top of the incline the mass is moving at 2.8 m/s. What is the work done by all non-conservative force in Joules?

Answers

Answer:

499.7 J

Explanation:

Since total mechanical energy is conserved,

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.

So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

Substituting the values of the variables into the equation, we have

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)²  + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)²  + W₂

0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s²  + W₂

907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s²  + W₂

907.38 kgm²/s² = 407.68 kgm²/s² + W₂

W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²

W₂ = 499.7 kgm²/s²

W₂ = 499.7 J

Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J

why does a spherometer have three legs?​

Answers

spherometer is a device used to measure curved in surface

it have 3 legs which form equivalent triangle.

geometry says that 3 point determine a plane that's why it have 3 legs


How much power does it take to lift 70.0 N to 5.0 m high in 5.00 s?

Answers

Answer:

Power = 70 W

Explanation:

Given that,

Force, F = 70 N

Height, h = 5 m

Time, t = 5 s

We need to find the power of the object. We know that,

Power = work done/time

Put all the values,

[tex]P=\dfrac{Fd}{t}\\\\P=\dfrac{70\times 5}{5}\\\\P=70\ W[/tex]

So, the required power is 70 W.


how to calculate sound of an echo ​

Answers

by an echo meter

please flw me and thank my answers

#Genius kudi

Calculate the current flowing when the voltage across is 35V and the resistance is 7ohms.​

Answers

Explanation:

V= IR

35=I×7

I=35/7

I=5amperes

pls give brainliest

You throw a football straight up. Air resistance can be neglected. When the football is 4.00 mm above where it left your hand, it is moving upward at 0.500 m/sm/s. What was the speed of the football when it left your hand

Answers

Answer:

u=8.868 m/s

Explanation:

The displacement of the ball is 4 meters

The final speed of the ball is 0.5 m/s

The initial speed of the ball is to be calculated. Using the equation of the rectilinear motion,

[tex]v^{2} =u^{2} +2as[/tex]

Plugging the values in the above expression,

[tex]\\0.5^{2} =u^{2} +2*(-9.8)*4\\\\u^{2} =78.4+0.25\\\\u^{2} =78.65\\\\u=8.868 m/s[/tex]

At 20 ◦C a copper wire has a resistance of 4×10−3 Ω and a temperature coefficient of resistivity of 3.9×10−3 (C◦)−1, its resistance at 100 ◦C is

A.
52.5 × 10-3 Ω
B.
5.25 × 10-3 Ω
C.
5.25 × 10-4 Ω
D.
5.25 × 10-2 Ω


E.
25.5 × 10-3 Ω

Answers

Answer:

[tex]R _{t} = R _{0}( \alpha t + 1) \\ = 4 \times {10}^{ - 3} (3.9 \times {10}^{ - 3} \times 20 + 1) \\ = 4 \times {10}^{ - 3} (1.078) \\ = 4.312 \times {10}^{ - 3} \: Ω[/tex]

Hydrogen carried in light phase​

Answers

Answer:

because it is helpful to human beings I think

In the light-dependent reactions, energy absorbed by sunlight is stored by two types of energy-carrier molecules: ATP and NADPH. ... The hydrogen ions are allowed to pass through the thylakoid membrane through an embedded protein complex called ATP synthase. This same protein generated ATP from ADP in the mitochondrion.

A lumberjack is trying to drag a small tree that he cut down. If the static
coefficient of friction of the tree on the ground is 0.5 and the tree weighs 430
N, what is the minimum amount of horizontal force that he will need to apply
so that the tree will start moving?
A. 215 N
B. 430 N
C. 365 N
D. 500 N

Answers

Answer:

A

Explanation:

weight of the tree =normal force

Horizontal force =coefficient of friction x Fnormal

0.5×430=215

A force of 3 newtons moves a 10 kilogram mass horizontally a distance of 3 meters. The mass does not slow down or speed up as it moves. Which of the following must be true?
a) 9 joules of kinetic energy were produced
b) 9 joules of gravitational potential energy were produced
c) 9 joules of heat energy were produced
d) 9 joules of kinetic energy and heat were produced

Answers

Answer:

9 joules of heat energy was produced

Explanation: there is no acceleration therefore its not a kinetic energy

Energy= force × distance

= 3×3

=9

PLEASE HELP ME WITH THIS ONE QUESTION
The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)

A) 2.81 eV

B) 3.89 eV

C) 2.10 eV

D) 2.78 eV

Answers

k = [tex] \dfrac{ (\dfrac{h}{ \lambda} )^{2} }{2m} [/tex]

k = (6.626×10-¹⁹/590 × 10-⁹ )^{2} /2 × 1.673 × 10-²⁷

k = (1.12 × 10-³⁰)^2/3.346×10-²⁷

k = 1.25 × 10-⁶⁰ /3.346×10-²⁷

k = 0

ldk why, my answer is coming this :(

Erica (37 kg ) and Danny (45 kg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is moving upward past her at 4.7 m/s . At that instant he grabs hold of her. What is their speed just after he grabs her?

Answers

Answer:

V = 2.58 m/s

Explanation:

Below is the calculation:

Given the weight of Erica = 37 kg

The weight of Danny = 45 kg

Danny's speed to move upward = 4.7 m/s

Use below formula to find the answer.

m1 * u1 = (m1+m2) * V

V = m1*u1 / (m1+m2)

Here, m1 = 45

u1 = 4.7

m1 = 45

m2 = 37

Now plug the values in formula:

V = m1*u1 / (m1+m2)

V = (45*4.7)/(45+37)

V = 2.58 m/s

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