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DaniilM [7]
3 years ago
13

A motorcyclist is traveling at 51.9 mph on a flat stretch of highway during a sudden rainstorm. The rain has reduced the coeffic

ient of static friction between the motorcycle's tires and the road to 0.076 when the motorcyclist slams on the brakes. (a) What is the minimum distance required to bring the motorcycle to a complete stop? (b) What would be the stopping distance if it were not raining and the coefficient of static friction were 0.580?

Physics
2 answers:
stira [4]3 years ago
8 0

Answer: a) 361.23m

b) 47.38m

Explanation:

Friction is opposite in direction to an applied force

Find the attached file for the solution

(b) The distance traveled with the friction of rain-free conditions is 47.38 m.

xz_007 [3.2K]3 years ago
7 0

Answer:

Part (a) 361m

Part (b) 47.4m

This problem involves involves the nonconservative frictional force. To solve this problem. The energy approacb will be used. The energy conservation is used to relate the work done by the frictional force over the required distance to the kinetic energy of the motorcycle before the brakes are applied.

Explanation:

The equation is

K1 + U1 + Wf = K2 + U2

Where U standa for potential energy which in this case is zero as the height of the motorcycle does not change over the distance of the application of the brakes to the tyres. The work of friction is negative.

The full solution can be found in the attachment below.

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A wave is a disturbance that carries
viktelen [127]

Answer:

(D) energy from one place to another

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2 years ago
Calcular la resistencia de una varilla de grafito de 170 cm de longitud y 60 mm2. Resistividad grafito 3,5 10-5 Ωm
ozzi

Answer:

R = 0.992 Ω

Explanation:

En esta pregunta, dada la información que contiene, debemos calcular la resistencia de la varilla de grafito.

Matemáticamente,

Resistencia = (resistividad * longitud) / Área De la pregunta;

Resistividad = 3,5 * 10 ^ -5 Ωm

longitud = 170 cm = 1,7 m

Área = 60 mm ^ 2 = 60/1000000 = 6 * 10 ^ -5 m ^ 2

Conectando estos valores a la ecuación anterior, tenemos;

Resistencia = (3.5 * 10 ^ -5 * 1.7) / (6 * 10 ^ -5) =

(3.5 * 1.7) / 6 = 0.992 Ω

3 0
3 years ago
One thousand grams of seawater would consist of how many grams of dissolved substances? 65 635 35 965
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<span>One thousand grams of seawater has 35 grams of dissolved substances ... on the average. While the salinity of the Earth's oceans and seas varies, the average salinity of seawater rests at 3.5%. Consider one liter or sea or ocean water. One liter has 1,000 milliliters (mL) in it. To find 3.5% of 1,000, we would multiply with the decimal place adjusted for percentages: 1000 x .035 = 35. Therefore, for every 1,000 mL of seawater, we will find 35 grams of (mostly) sodium chloride, otherwise known as salt.</span>
7 0
2 years ago
Read 2 more answers
Air pollution is an example of a. The underground economy. B. An external shock. C. A nonmarket activity. D. A negative external
devlian [24]

Answer:

D. A negative externality.

Explanation:

Particulate pollution is a form of pollution that is responsible for the degradation of the environment.

Particulate matter is also referred to as particle pollution or atmospheric aerosol particles and it can be defined as a complex microscopic mixture of liquid droplets and solid particles that are suspended in air.

An externality is typically an unwarranted cost or benefit by a manufacturer or producer of goods and services that affects a third party.

In Economics, an externality could either be positive or negative depending on its effect on a third party.

A negative externality arises when the production or consumption of a finished product or service has negative impact (cost) on a third party.

In conclusion, air pollution is an example of a negative externality because it causes harm to a third party.

3 0
2 years ago
Two Carnot air conditioners, A and B, are removing heat from different rooms. The outside temperature is the same for both rooms
lakkis [162]

Answer:

(a)  333.77 J

(b) 237.85 J

(c) 4763.77 J

(d) 4667.85 J

Explanation:

Temperature of source, TH = 314 K

Temperature of A, Tc = 292 K

Temperature of B, Tc' = 298 K

heat taken out, Qc = 4430 J

Let the heat deposited outside is QH and QH' by A and B respectively.

\frac{Q_H}{Q_c}=\frac{T_H}{T_c}\\\\Q_H = \frac{4430\times314}{292}=4763.77 J

Now

\frac{Q_H'}{Q_c}=\frac{T_H}{T_c'}\\\\Q_H' = \frac{4430\times314}{298}=4667.85 J

(a) Work done for A

W = QH - QC = 4763.77 - 4430 = 333.77 J

(b) Work done for B

W' = QH' - Qc = 4667.85 - 4430 = 237.85 J

(c) QH = 4763.77 J

(d) QH' = 4667.85 J

4 0
2 years ago
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