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kolbaska11 [484]

3 years ago

6

You are driving on the highway at a speed of 40 m/s (which is over the speed limit) when you notice a cop in front of you. To av

oid a ticket, you press on the brake and slow to a speed of 30 m/s over the course of 5 seconds. What is the acceleration of the car? SHOW WORK OR GET REPORTED(have 5 accounts)
What is your car's initial velocity?

What is your car's final velocity?

How long does it take the car to slow down?

Write the equation you will use to solve this problem.

What is the acceleration of your vehicle?
+ 2.0 m/s^2
- 2.0 m/s^2
+ 8.0 m/s^2
- 6.0 m/s^2

1 answer:

ladessa [460]3 years ago

3 0

Answer:

What is the acceleration of your vehicle?

+ 2.0 m/s^2 ✔

- 2.0 m/s^2

+ 8.0 m/s^2

- 6.0 m/s^2

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A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm

Karo-lina-s [1.5K]

Answer:

the thermistor temperature = $325.68 \ ^0 \ C$

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

$T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K$

Resistance of the thermistor $R_1 =$ 20,000 ohms

Material constant $\beta$ = 3650

Resistance of the thermistor $R_2$ = 500 ohms

Using the equation :

$R_1 = R_2 \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{R_1}{ R_2} = \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

Taking log of both sides

$In \ \frac{R_1}{ R_2} = In \ \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{1}{T_2} = \frac{1}{T_1} - \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}$

${T_2} = \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}$

Replacing our values into the above equation :

${T_2} = \frac{3650*373}{3650 - In (\frac{20000}{500})373}$

${T_2} = \frac{1361450}{3650 - 3.6888*373}$

${T_2} = \frac{1361450}{3650 - 1375.92}$

${T_2} = \frac{1361450}{2274.08}$

${T_2} = 598.68 \ K$

${T_2} = 325.68 \ ^0 \ C$

Thus, the thermistor temperature = $325.68 \ ^0 \ C$

4 0

3 years ago

A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i

Hoochie [10]

The acceleration of the box is approximately $1 m/s^2$

Explanation:

According to Newton's second law of motion, the net force acting on the box is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m = 12.0 kg is the mass of the box

a is the acceleration

The net force can be written as

$\sum F = F_a - F_f$

where

$F_a = 20 N$ is the applied forward force

$F_f=9.0 N$ is the friction force

Combining the two equations,

$F_a-F_f=ma$

And solving for the acceleration,

$a=\frac{F_a-F_f}{m}=\frac{20-9}{12}=0.9 m/s^2\sim 1 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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