At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to (a) 2 times that of the Celsius and (b) 1/4 times that of the Celsius

Answers

Answer 1

Answer:

(a) F = 320

(b) = F = -5.1625

Explanation:

The formula that converts degree Celsius (C) to degree Fahrenheit (F) is:

F = 1.8C + 32

Solving (a): F = 2C

Substitute 2C for F in the above equation

F = 1.8C + 32

2C = 1.8C + 32

Collect like terms

2C - 1.8C = 32

0.2C = 32

Multiply both sides by 5

5 * 0.2C = 32 * 5

C = 160

Recall that F = 2C

F = 2 * 160

F = 320

Solving (b): F = ¼C

Substitute ¼C for F in the above formula

F = 1.8C + 32

¼C = 1.8C + 32

Convert fraction to decimal

0.25C = 1.8C + 32

Collect like terms

0.25C - 1.8C = 32

-1.55C = 32

Divide both sides by -1.55

C = 32/(-1.55)

C = -32/1.55

C = -20.65

Recall that: F = ¼C

F = -¼ * 20.65

F = -5.1625


Related Questions

Need help please!!!!!!!!!!!!!!!!!

Answers

Answer:

3 times

Explanation:

F=GMm/^2

if M is tripled,

F'=G3Mm/r^2

→F'=3F

Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 480 N/C. If the particles are free to move, what are their speeds (in m/s) after 48.4 ns

Answers

Answer:

The speed of electron is 4.08 x 10 m/sThe speed of proton is 2,225.82 m/s

Explanation:

Given;

electric field, E = 480 N/C

mass of electron, Me = 9.11 x 10⁻³¹ kg

mass of proton, Mp = 1.67 x 10⁻²⁷ kg

time of motion, t = 48.4 ns = 48.4 x 10⁻⁹ s

initial velocity of the particles, u = 0 (initially at rest)

let the speed of each particle after 48.4 ns = v

the magnitude of charge of the particles, q = 1.6 x 10⁻¹⁹ C

The force experienced by each particle is calculated as;

F = Eq

F = (480 N/C) x (1.6 x 10⁻¹⁹ C)

F = 7.68 x 10⁻¹⁷ N

The speed of each particle after 48.4 ns is calculated as;

[tex]F = ma\\\\F = \frac{m(v-u)}{t} \\\\F = \frac{m(v-0)}{t}\\\\F = \frac{mv}{t} \\\\mv = Ft\\\\v = \frac{Ft}{m} \\\\For \ electron;\\\\v_e = \frac{Ft}{m_e} \\\\v_e = \frac{7.68 \times 10^{-17} \ \times \ 48.4 \times 10^{-9}}{9.11 \times 10^{-31}} \\\\v_e = 4.08 \times 10^6 \ m/s[/tex]

[tex]For \ proton;\\\\v_p = \frac{Ft}{m_p} \\\\v_p = \frac{7.68 \times 10^{-17} \ \times \ 48.4 \times 10^{-9}}{1.67 \times 10^{-27}} \\\\v_p = 2,225 .82 \ m/s[/tex]

a man weighs 600 n while on the surface of earth. if he is transported to the planet mythos which has the same mass as earth but a radius tjat os five times larger than earths , his weight would be

Answers

Answer:

24N

Explanation:

Calculation for what his weight would be

Let his WEIGHT on the surface of earth be 600 and Let the RADIUS be 25 which is (5^²) because we were told that the radius is FIVE TIMES larger than earths

Now let calculate his weight using this formula

Weight=Weight on Earth/Radius

Let plug in the formula

Weight=600N/5^²

Weight=600N/25

Weight=24N

Therefore his weight would be 24N

Chirag, a champion weightlifter, raises 259 kg of weights a distance of 2.40 m. How much work is done by Brutus lifting the weights?

622 J
6090 J
2540 J
0 J

How much work is done by Chirag holding the weights above his head?
622 J
6090 J
2540 J
0 J

Answers

1.

6090 J

2.

622 J

I bet so~

The answers are
1.6090j
2.622j

A weight lifter lifts a 360-N set of weights from ground level to a position over his head, a vertical distance of 1.95 m. How much work does the weight lifter do, assuming he moves the weights at constant speed

Answers

Answer:

W= 702 J

Explanation:

Given that,

A weight lifter lifts a 360-N set of weights from ground level to a position over his head, a vertical distance of 1.95 m.

We need to find the work the weight lifter do. Work done by an object is given by the formula as follows :

W = Fd

Putting all the values,

W = 360 N × 1.95 m

= 702 J

So, the required work done is 702 J.

Which layer of the atmosphere is most likely to protect life on Earth from ultraviolet radiation

Answers

Answer:

ozone layer

Explanation:

The ozone layer forms a thin shield in the upper atmosphere protecting life on earth from UV rays. Ozone is a naturally occurring gas that is found in two layers of the atmosphere

Popeye, of mass 75.0 kg, has just downed a can of spinach. He accelerates quickly and stops Bluto, of mass 720 kg (Bluto is very dense), who is charging in at 15.0 m/s. What was Popeye's speed

Answers

Answer:

Popeye's speed is 144 m/s.

Explanation:

Given;

mass of Popeye, m₁ = 75 kg

mass of Bluto, m₂ = 720 kg

velocity of Bluto, v₂ = 15 m/a

velocity of Popeye, v₁ = ?

Apply the principle of conservation of linear momentum;

m₁v₁ = m₂v₂

v₁ = (m₂v₂) / m₁

[tex]v_1 = \frac{720\ \times \ 15}{75} \\\\v_1 = 144 \ m/s[/tex]

Therefore, Popeye's speed is 144 m/s.

A toy car rolls down a straight inclined track and its speed increases steadily. At the moment when the car reaches the flat surface at the base of the track, its speed remains constant. Which statement best explains why the speed of the car remains the same when it reaches the flat surface

Answers

Answer:

Due to same gravity.

Explanation:

The speed of the car remains the same when it reaches on the flat surface because the gravity is same at every step of flat surface while on the other hand, the speed of car speed increases steadily on the inclined plane due to gravity of the earth. So we can say that speed of the car remains the same on the flat surface due to same gravitational force.

all 4 questions plz ( pls hurry)

Answers

1. C
2. B
3. A
4. B

Explanation

Write a paragraph describing Newton's 2nd Law in YOUR OWN WORDS.

Answers

Explanation:

if u apply more force to an object, it accelerates at a higher rate

A 25 kg box on a horizontal frictionless surface is moving to the right at a speed of 5.0 m s. The box hits and remains attached to one end of a spring of negligible mass whose other end is attached to a wall. As a result, the spring compresses a maximum distance of 0.60 m. (a) i. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maximum compression. Indicate whether the work done by the spring is positive, negative, or zero.

Answers

Answer:

Explanation:

Work done by the spring is negative .

Work done by force F creating displacement  d is given by the following expression .

Work = F x d

Both force and displacement are vector quantity .

When direction of force and direction of displacement is same , work is positive . When direction of force and direction of displacement is opposite , work is negative .

When spring is compressed , it exerts a restoring or opposing force in a direction opposite to the direction of displacement of box . Hence here force is opposite to displacement . Restoring force acts opposite to displacement . Hence work done by spring on box is negative .

An unstrained horizontal spring has a length of 0.26 m and a spring constant of 180 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.039 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.

Answers

Answer:

a)

two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)

b)

both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

Explanation:

Given that;

L = 0.26 m

k = 180 N/m

x = 0.039 m

a)

we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.

b)

Spring force F = kx

F = 180 × 0.039

F = 7.02 N

Now, Electrostatic force F = Keq²/r²

where r = L + x = ( 0.26 + 0.039 )

we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2

so from the equation; F = Keq²/r²

Fr² = Keq²

q = √ ( Fr² / Ke )

we substitute

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q =  √ (0.627595 / 9×10⁹)

q = √(6.97 × 10⁻¹¹)

q = 8.35 × 10⁻⁶ C

Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

2 objects have a total momentum of 400kg m/s, they collide. Object A’s mass is5kg & object B’s mass is 11kg. After the collision Object B is moving at 15m/s.What is the velocity of Object A AFTER the collision?

Answers

Answer:

Explanation:

We shall apply law of conservation of momentum .

Momentum before collision = momentum after collision .

Momentum before collision = 400 kg m/s

Momentum after collision = 5  x v + 11 x 15

where v is velocity of A after the collision .

5  x v + 11 x 15 = 400

5 v = 400 - 165

5v = 235

v = 47 m /s .

1.
In the notation, 1s2, the number 1 represents the ________ __________ where electrons are located.
energy level
outer shell
2.
The four parts, or blocks, on the periodic table are _____, _____, d, and f.
s, p
a, b
3.
The 2 in 1s2 represents the number of __________ located in that energy level.
electrons
protons

Answers

Answer:

energy level

s,p

electrons

Explanation:

I already done this trust me

In the notation 1 s² number 1 represents energy levels where electrons are located.The 4 parts on periodic table are s,p,d and f blocks.The 2 in 1 s² represents number of electrons in energy level.

What is an energy level?

Electrons present in an atom revolve in different orbits which are stationary states and are also called as energy levels. The energy levels are numbered as integers which are also called as principal quantum numbers.

Energy of the stationary state is given as E= -R

1/n² where R

is the Rydberg's constant. When an electron is excited, and it moves from lower to higher energy levels there is absorption of energy, while when it moves from higher energy level to lower energy level it radiates or gives out energy in the form of radiation.

They can also be defined as the distances between electron and nucleus of an atom . Electrons present in K energy level have least energy .Energy level diagrams are studied to understand nature of bonding , placement of electrons in orbits and and elemental behavior under certain conditions.

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Use the following information about earth and its moon to determine the distance between them

Answers

answer is the SECOND one on Edge

3.8 x 10^8m

the chart would’ve been helpful though.

Answer:

B

Explanation:

edge2020

Explain why gravity is the only force acting on a falling ball

Answers

Answer:

Gravity is the only force acting on a falling ball because of free fall. Since gravity is unbalanced it accelerates an object and the velocity increases as an object falls due to gravity pushing it. The acceleration due to gravity is 9.8 m/s^2.

what is phyical quantity​

Answers

Answer:

physical quantity

Explanation:

All quantity in term of which laws of physics are described and can be measured is called physical quantity

Estimate the weight of a 1000kg car that is accelerating at 3 m/s/s.

Answers

Answer:

W = 9800  N

Explanation:

Given that,

Mass of a car, m = 1000 kg

Acceleration of the car, a = 3 m/s²

We need to find the weight of the car. Weight of an object is given by the product of mass and acceleration due to gravity on the Earth.

W = mg

Put all the values,

W = 1000 kg × 9.8 m/s²

= 9800  N

So, the weight of the car is 9800  N.

Which scenario is best illustrates the concept of illusory correlation?
A. A person insists that there is no way to determine whether or not
any two events are actually correlated.
B. A person believes crossing his fingers brings good luck despite
studies demonstrating that it does not.
O C. A person argues that the mood swings he often has on rainy days
prove that weather affects human emotions.
O D. A person criticizes a scientific study for reaching a definitive
conclusion about data based on a weak correlation.

Answers

Answer:

A person believes crossing his fingers brings good luck despite studies demonstrating that it does not.

Explanation:

Hope it helps!

A human is lifting their arm against the force of gravity using only the biceps brachii. The muscle inserts 0.04 m from the joint axis of rotation, and generates a force of 55 N. The mass of the forearm & hand is 1.84 kg, the center of mass is located 0.11 m from the joint axis, and the moment of inertia is 3.4 kg m2. Assume the arm is at the right angle.

Required:
a. What is the net torque at the elbow joint?
b. What is the angular acceleration of the forearm?
c, What force would the biceps need to generate to keep the hand steady?

Answers

Answer:

a) ∑ tau = - 1.9965 N m , b) α = 0.587 rad / s²,  c) F₁ = 5.09 N

Explanation:

a) the torque is given by the expression

             Σ τ = Σ F x r

for this case we assume that the counterclockwise rotations are positive

             

In the problem, the right arm F₁ = 55 N applied at a distance x₁= 0.04m and the weight of the forearm mass W = 1.84 kg applied to x₂ = 0.11 m gives us two forces, this force has to create a rotation counterclockwise thereby creating positive torque

           ∑ tau = -F₁ x₁ + W x₂

           ∑ tau = - 55 0.04 + 1.85 0.11

           ∑ tau = - 1.9965 N m

b) the angular acceleration can be obtained from Newton's second law for rotations

          τ = I α

           α = τ/I

           α = 1.9965 / 3.4

           α = 0.587 rad / s²

c) what is the strength of the biceps to maintain balance

           ∑ τ = 0

           -F₁ x₁ + W x₂ = 0

            F₁ = [tex]\frac{x_{2} }{ x_{1} }[/tex]  W

            F₁ = [tex]\frac{0.11}{0.04}[/tex]  1.85

            F₁ = 5.09 N

A He-Ne laser produces 633 nm light.
When it passes through a diffraction
grating with d = 5.55 x 10^-6 m, what is
the angle of its third order (m = 3)
maximum?

Answers

Answer:

20.0086

Explanation:sin^-1(3×633×10^-9)÷5.55×10^-6

Answer:

20.0086

Explanation:

Acellus

The Moon has a mass of 7.36 1022 kg and a radius of 1.74 106 m. (a) What is the acceleration due to gravity on the Moon

Answers

Answer:

1.622m/s²

Explanation:

Given the following

mass of 7.36 × 10²² kg

Radius r = 1.74×10^6m

Gravitational constant G = 6.67×10^-11

Acceleration due to gravity us expressed as:

g = GM/r²

Substitute the given values into the formula

g = 6.67×10^-11×7.36 × 10²²/(1.74×10^6)²

g = 49.0912×10¹¹/3.0276×10¹²

g = 16.22×10^{11-12}

g = 16.22×10^-1

g = 1.622m/s²

Hence the acceleration due to gravity on the Moon is 1.622m/s²

The 1360-kg car and the 2100-kg tow truck are stationary. The muddy surface on which the car s tires rest exerts negligible friction forces on them. What is the tension in the tow cable

Answers

Answer:

5.91 kN

Explanation:

The schematic view of the free body diagram of the car is shown in the image attached below.

Let's first calculate the weight of the car.

W = mg

W = 1360 × 9.81

W = 13341.6 N

W = 13.3416 kN

From the image, Using the equilibrium equations, we have:

[tex]\sum F_x = 0\\[/tex]

T cos (18° - 10°) - 13.3416 sin 26° = 0

T cos 8° - 5.84857 = 0

T cos 8° = 5.84857

[tex]T = \dfrac{5.84857}{cos \ 8^0}[/tex]

[tex]T = \dfrac{5.84857}{0.9903}[/tex]

T = 5.905856811 N

T ≅ 5.91 kN


What was the main cause of the shift to a postindustrial society?

A. Citizens started to prefer manual labor to jobs that required providing a service.

B. The economic situation allowed people to work from home instead of going to work.

C. The focus became the control of information instead of traditional manufacturing.

D. People started to purchase computers and cell phones in mass quantities.

Answers

Answer: i thnk its B

Explanation:

what does correlation mean

Answers

a mutual relationship or connection between two or more things.

A ball is thrown horizontally from the top of a 73.9-m building and lands 183 m from the base of the building. Ignore air resistance. How long is the ball in the air

Answers

Answer:

t = 3.88 seconds

Explanation:

Given that,

A ball is thrown horizontally from the top of a 73.9-m building and lands 183 m from the base of the building.

We need to find the time for which the ball is in air. Let the time be t.

[tex]y=y_o+ut+\dfrac{1}{2}at^2[/tex]

Here, a = -g, [tex]y_o=73.9\ m[/tex] and y = 0

u is initial velocity, u = 0

[tex]y_o=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2y_o}{g}}\\\\t=\sqrt{\dfrac{2\times 73.9}{9.8}}\\\\t=3.88\ s[/tex]

So, the ball is in the air for 3.88 seconds.

The average velocity of a tennis ball is measured during serving and found to be 216 km/hr.
Calculate the time elapsed between hitting the ball and meeting the ground if the distance covered by the ball during that interval is 20m​
PLEASE ANSWER WITH EXPLANATION :)

Answers

Answer:

The time between hitting the ball and meeting the ground is [tex]t=0.33\: s[/tex]

Explanation:

We have a semi parabolic motion here. We know that the initial speed 216 km/h or 60 m/s in the x-direction.

So we can use the following equation:

[tex]v=\frac{\Delta x}{\Delta t}[/tex]

[tex]60=\frac{20}{t}[/tex]

[tex]t=\frac{20}{60}[/tex]

[tex]t=0.33\: s[/tex]

hope it helps you!

The period of a pendulum is directly proportional to the mass of the bob.

true or false.

Answers

Answer:

false

Explanation:

i just took the test <3

The period of a pendulum is not directly proportional to the mass of the bob. The period of pendulum is independent of the mass of the bob. Thus, the given statement is false.

What is the period of pendulum?

A simple pendulum is the one which consists of a small metal ball called as the bob or a mass which is suspended from a fixed point by a long piece of thread such that the bob is free to swing back and forth from its mean position under the influence of the gravity.

The time period of a simple pendulum is the time taken by it to complete one oscillation. The formula for the time period or period (T) of a simple pendulum is T = 2π (√L/g), where L is the length of the pendulum thread and g is the acceleration due to gravity.

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A 0.877-kg ball has an elastic, head-on collision with a second ball that is initially at rest. The second ball moves off with half the original speed of the first ball. (a) What is the mass of the second ball

Answers

Answer: the mass of the second ball is 2.631 kg

Explanation:

Given that;

m1 = 0.877 kg

Initial velocity = V0

Initial momentum = m1 × V0

final velocity of m1 is u1, final velocity of m2 is u2 = v0/2

now final momentum = m1 × u1 + m2 × u2

using momentum conservation;

m1×V0 = m1×u1 + m2×v0/2

m1×(v0 - u1) = m2×V0/2 ----- let this be equation 1

Now, for elastic collision;

m1×v0²/2 = m1×u1²/2 + m2×(v0/2)²/2

m1×(v0² - u1²) = m2×(v0/2)² --------- let this be equation 2

now; equation 2 / equation 1

: V0 + u1 = v0/2

2V0 + 2u1 = V0

2u1 = V0 - 2V0

u1 = -V0/2

now we insert in equ 1

 m1×3V0/2= m2×V0/2

m1 × 3 = m2

m2 = 0.877 × 3

m2 = 2.631 kg

Therefore, the mass of the second ball is 2.631 kg

A dragster going at 15 m/s increases its velocity to 25 m/s in in 2 seconds. What is its acceleration?

Answers

Answer:

5m/s²

Explanation:

Given parameters:

Initial velocity  = 15m/s

Final velocity  = 25m/s

Time taken  = 2s

Unknown:

Acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time taken. It is mathematically expressed as;

      Acceleration  = [tex]\frac{Final velocity - Initial velocity }{Time taken }[/tex]  

    Acceleration  = [tex]\frac{25 - 15}{2}[/tex]   = 5m/s²

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