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3 years ago

What average net force is required to stop a 1950 kg car in 10.5 s if it’s initially traveling at 28m/s

Physics

1 answer:

nika2105 [10]3 years ago

7 0

Answer:

An average net force of 5200 N is needed to stop the car

Explanation:

Cinematics and Dynamics

Cinematics describes the variables involved in the movement without dealing with its causes. There are four main concepts in cinematics: Velocity (or its scalar equivalent, the speed), acceleration, time, and displacement (or the scalar equivalent, distance).

The acceleration can be calculated by:

$\displaystyle a=\frac{v_f-v_o}{t}$

The initial speed is vo=28 m/s, it stops (vf=0) in t=10.5 seconds, thus the acceleration is:

$\displaystyle a=\frac{0-28}{10.5}$

$a = -2.67~m/s^2$

The acceleration is negative because the car loses speed.

Knowing the mass of the car m=1950 Kg, we can calculate the net force required to stop the car by using the formula:

F = m.a =1950*2.67

F = 5200 N

An average net force of 5200 N is needed to stop the car

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The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization

ANEK [815]

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

$ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]$

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

$ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg$

Therefore, the vapor pressure at 60.6°C is 330.89 mmHg

6 0

3 years ago

Read 2 more answers

A car Excel accelerates from the rest to a velocity of 5 m in four seconds what is the average acceleration

MrRa [10]

Average acceleration: velocity/time

5/4= 1.25 m/s² Acceleration

Hope I helped :)

5 0

3 years ago

Jeff is part of a trekking team. As he climbs a hill, he drops his water bottle, which has a mass of 0.25 kilograms.

yulyashka [42]

The vertical height from where Jeff dropped the bottle is 10m.

<h3>What is free falling?</h3>

When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy(P.E) of

U =mgh

When the object strikes the ground, all the potential energy converted into kinetic energy.

k.E = 1/2mv²

where v is the speed just before hitting the ground.

From energy conservation principle, initial and final mechanical energy are equal.

mgh = 1/2mv²

Given is Jeff climbs a hill. He drops his water bottle, which has a mass of 0.25 kilograms. The bottle slides down the hill and is moving at a velocity of 14 meters/second the instant it hits the ground.

The height from which bottle could fall is

h = v²/2g

Substitute the values, we get

h = (14)²/(2x9.81)

h =9.99 m

Thus, vertical height from where Jeff dropped the bottle is approximately 10m.

Learn more about free falling.

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8 0

2 years ago

What is the root mean square speed of an As4 particle as it is sublimed? (Assume at high temperatures arsenic acts like an ideal

Norma-Jean [14]

Answer: $v_{rms} =$ 273m/s

Explanation: Root Mean Square Speed of an atom or molecule is the speed of a particle in a gas. It is the average speed a particle in a gas can have.

It can be calculated:

$\frac{1}{2}mv^{2} = \frac{3}{2} k_{B}T$