How long will it take a car to accelerate from 15.2 to 23.5 m/s if the car has an average acceleration of 3.2 m/s?

Answers

Answer 1
Answer:

It will take a car, 2.59 s to accelerate from 15.2 to 23.5 m/s.

What is Speed?

speed is described as. the pace at which an object's location changes in any direction. Speed is defined as the distance traveled divided by the travel time. Speed is a scalar quantity because it just has a direction and no magnitude.

Given, the car has an average acceleration of 3.2 m/s².

To solve this problem, we can use the following kinematic equation:

v = u +at

where:

v is the final velocity (23.5 m/s)

u is the initial velocity (15.2 m/s)

a is the acceleration (3.2 m/s^2)

t is the time

We can rearrange this equation to solve for t:

t = (v -u)/a

substituting the values we have:

t = (23.5 - 15.2 ) / 3.2

t = 2.59375 seconds

Therefore, it will take approximately 2.59 seconds for the car to accelerate from 15.2 m/s to 23.5 m/s with an average acceleration of 3.2 m/s².

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Answer 2
Answer:

Hello!

How long will it take a car to accelerate from 15.2 m/s to 23.5 m/s if the car has an average acceleration of 3.2 m/s² ?

We have the following data:

Vf (final velocity) = 23.5 m/s

Vi (initial velocity) = 15.2 m/s

ΔV  (speed interval)  = Vf - Vi → ΔV  = 23.5 - 15.2 → ΔV  = 8.3 m/s

ΔT (time interval) = ? (in s)

a (average acceleration) = 3.2 m/s²

Formula:

a = \frac{\Delta{V}}{\Delta{T^}}

Solving:  

a = \frac{\Delta{V}}{\Delta{T^}}

3.2 = \frac{8.3}{\Delta{T^}}

\Delta{T^} = (8.3)/(3.2)

\Delta{T^} = 2.59375 \to \boxed{\boxed{\Delta{T^} \approx 2.6\:s}}\:\:\:\:\:\:\bf\green{\checkmark}

Answer:  

The car will take approximately 2.6 seconds to accelerate

____________________________________  

I Hope this helps, greetings ... Dexteright02! =)


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Answers

Answer:

9 answer hai brown

You use a 1200-watt hair dryer for 10 minutes each day. a. How many minutes do you use the hair dryer in a month? (Assume there are 30 days in the month.) b. How many hours do you use the hair dryer in a month? c. What is the power of the hair dryer in kilowatts? d. How many kilowatt-hours of electricity does the hair dryer use in a month? e. If your town charges $0.15/kWh, what is the cost to use the hair dryer for a month?

Answers

Data;

  • Power = 1200W
  • Time/day = 10 mins

Power

The is the rate of energy used to time. It is the measure of expedience of energy.

a). How many minutes is the dryer used in a month.

Assuming there 30 days in a month;

1 day = 10 mins\n30 days = x mins\nx = 30 * 10 \nx = 300 mins

This shows that the dryer is used 300 minutes in a month.

b). How many hours is the dryer used in a month.

To solve this problem, we simply need to convert the value of (a) from minutes to hours.

1hr = 60 mins\ny hr = 300 mins\ny = (300)/(60) \ny = 5 hours

The dryer was used for a total of 5 hours in a month.

c). The power of the dryer in  kilowatt.

To convert the power in watt to kilowatt, we divide by 1000.

P = 1200W\nP = (1200)/(1000) \nP = 1.2kw

The power in kilowatt is 1.2kw

d). How many kilowatt/hr of electricity used in a month.

1.2 * 5 = 6kw/h

The dryer consumes 6kw/h in a month.

e). If the town charges $0.15/kwh, the cost in a month

\$0.15 = 1hr\nx= 6hr\nx = 6 * 0.15\nx = 0.9

The cost of the dryer in a month is $0.9.

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300 minuts
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$0.9

A place kicker must kick a football from a point 36.0 m from the uprights. As a result of the kick, the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with an initial speed of 20 m/s at an angle of 53° to the horizontal. By how much does the ball clear or fall short of clearing the crossbar?

Answers

Final answer:

By utilizing principles of projectile motion, it is found that the football clears the crossbar by approximately 10.75 meters.

Explanation:

To determine by how much the ball clears or falls short of clearing the crossbar, we need to use the physics principles of projectile motion. The maximum height 'h' of the football can be given by equation of motion: h = (v²sin²θ) / (2g), where 'v' is the initial velocity, 'g' is the acceleration due to gravity and 'θ' is the angle of projection.

Substituting the given values: h = [(20)²sin²53°] / (2*9.8) ≈ 13.8 m above the ground when it was kicked. However, the football was kicked from ground level, so we need to subtract the height of the crossbar from this value, which is 3.05 m. Thus, the ball clears the crossbar by approximately: 13.8 - 3.05 = 10.75 m.

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Name the scientist that showed copernicus's theory of a moving earth reasonable?

Answers

The name of the Scientist that showed Copernicus theory of a moving earth reasonable was Galileo. Hope this helps you.
The scientist was Galileo Galelei

Please answer the circled questions, I will give you brainliest

Answers

Answer:

yo so if nobody helps ill gladly take the brainliest

Explanation:

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.5 x 10⁻⁷ C/m², and the plates are separated by a distance of 1.7 x 10⁻² m. How fast is the electron moving just before it reaches the positive plate?

Answers

The speed of the electron before reaching the positive plate is 1.30 * 10^(7)\ m / s

Explanation:

As per Gauss law of electro statistics, the electric field generated by a capacitor is directly proportional to the surface charge density of the plate and inversely proportional to the dielectric constant. In simple words, the electric field is proportional to the surface charge density.  So,  

    \text {Electric field}=(\sigma)/(\varepsilon_(0))

And then from the second law of motion, F=m * acceleration

So acceleration exerted by the electrons will be directly proportional to the force exerted on them and inversely proportional to the mass of the electron.

        Acceleration =(F)/(m)

Since force is also calculated as product of charge with electric field in electrostatic force,

       \text {Acceleration}=(q E)/(m)=(q \sigma)/(m \varepsilon_(0))

So, the charge of electronq=1.6 * 10^(-19)\ \mathrm{C}, \sigma=\text { Charge per unit area }=2.5 * 10^(-7)\ \mathrm{C} / \mathrm{m}^(2)

m is the mass of electron which is equal to 9.11 * 10^(-31)\ \mathrm{kg}

\varepsilon_(0)=8.85 * 10^(-12)\ \mathrm{Nm}^(2) \mathrm{C}^(-2)

Then,

    \text { Acceleration }=(1.6 * 2.5 * 10^(-19) * 10^(-7))/(9.11 * 8.85 * 10^(-31) * 10^(-12))=(4 * 10^(-19-7))/(80.62 * 10^(-31-12))

   \text { Acceleration }=0.0496 * 10^(-19-7+31+12)=0.0496 * 10^(17)\ \mathrm{m} / \mathrm{s}^(2)

So the acceleration of the electron in the capacitor will be 4.96 * 10^(15) m / s^(2)

Then, the velocity can be observed from the third equation of motion.

    v^(2)=u^(2)+2 a s

As u = 0 and s is the distance of separation between two plates.

   \begin{array}{c}v^(2)=0+\left(2 * 4.96 * 10^(15) * 1.7 * 10^(-2)\right) \nv^(2)=16.864 * 10^(15-2)=16.864 * 10^(13)=1.684 * 10^(14)\end{array}

Thus, v=\sqrt{\left(1.68 * 10^(14)\right)}=1.30 * 10^(7)\ m/s

So, the speed of the electron before reaching the positive plate is 1.30 * 10^(7) \mathrm{m} / \mathrm{s}.