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3 years ago

Which factor caused higher oil prices to directly lead to inflation?

Physics

2 answers:

mr Goodwill [35]3 years ago

8 0

Answer: B, Companies passed on production and transportation costs to consumers

Explanation:

Natali [406]3 years ago

8 0

Answer:

B on Edge 2020

Explanation:

You might be interested in

Wilbur ran 1-kilometer. Then he ran 500 meters. How many meters did Wilbur run all together?

Blababa [14]

The answer is B 1,500 meters since 1 kilometer to meters is 1,000 and u add the 500

8 0

3 years ago

1. A step-up transformer increases 15.7V to 110V. What is the current in the secondary as compared to the primary? Assume 100 pe

Serjik [45]

1. $I_2 = 0.14 I_1$

Explanation:

We have:

$V_1 = 15.7 V$ voltage in the primary coil

$V_2 = 110 V$ voltage in the secondary coil

The efficiency of the transformer is 100%: this means that the power in the primary coil and in the secondary coil are equal

$P_1 = P_2\\V_1 I_1 = V_2 I_2$

where I1 and I2 are the currents in the two coils. Re-arranging the equation, we find

$\frac{I_2}{I_1}=\frac{V_1}{V_2}=\frac{15.7 V}{110 V}=0.14$

which means that the current in the secondary coil is 14% of the value of the current in the primary coil.

2. 5.7 V

We can solve the problem by using the transformer equation:

$\frac{N_p}{N_s}=\frac{V_p}{V_s}$

where:

Np = 400 is the number of turns in the primary coil

Ns = 19 is the number of turns in the secondary coil

Vp = 120 V is the voltage in the primary coil

Vs = ? is the voltage in the secondary coil

Re-arranging the formula and substituting the numbers, we find:

$V_s = V_p \frac{N_s}{N_p}=(120 V)\frac{19}{400}=5.7 V$

5 0

3 years ago

by how many times occur in the force of attraction between two bodies change when the distance between then is reduced to one th

xenn [34]

Answer:

The force is now 9 times the original force

Explanation:

Coulomb's Law

The electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Coulomb's formula is:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the particles' charge

d= The distance between the particles

Suppose the distance is reduced to d'=d/3, the new force F' is:

$\displaystyle F'=k\frac{q_1q_2}{\left(\frac{d}{3}\right)^2}$

$\displaystyle F'=k\frac{q_1q_2}{\frac{d^2}{9}}$

$\displaystyle F'=9k\frac{q_1q_2}{d^2}$

$\displaystyle F'=9F$

The force is now 9 times the original force

8 0

2 years ago

1000 millicoulombs of charge passes through a point. The amount of current passing through the point is

Rzqust [24]

The amount of current passing through the point is 1 A

The amount of current passing through the point can be calculated using the formula below.

⇒ Formula:

Q = i/t......................... Equation 1

⇒ Where:

Q = Charge
i = current
t = time.

⇒ Make "i" the subject of the equation.

i = Qt....................... Equation 2

From the question,

⇒ Given:

Q = 1000 millicoulombs = 1 coulombs
t = 1 seconds. (Assuming the time is 1 seconds)

⇒ Substitute these values into equation 2

i = 1/1
i = 1 A.

Hence, The amount of current passing through the point is 1 A.

Learn more about charges here: brainly.com/question/4158552

3 0

2 years ago

The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the

tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.

A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.

B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

6 0

3 years ago