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2 years ago

10

In 1991 four English teenagers built an eletric car that could attain a speed of 30.0m/s. Suppose it takes 8.0s for this car to

accelerate from 18.0m/s to 30.0m/s. What is the magnitude of the car's acceleration?

2 answers:

ehidna [41]2 years ago

8 0

Answer:

Acceleration of the car, $a=1.5\ m/s^2$

Explanation:

It is given that,

Initial speed of the car, u = 18 m/s

Final speed of the car, v = 30 m/s

Time taken, t = 8 s

We need to find the acceleration of the car. Let it is given by a. It can be calculated using the first equation of motion.

$v=u+at$

Where

a is the acceleration of the car.

$a=\dfrac{v-u}{t}$

$a=\dfrac{30-18}{8}$

$a=1.5\ m/s^2$

So, the magnitude of the car's acceleration is $1.5\ m/s^2$ . Hence, this is the required solution.

ivanzaharov [21]2 years ago

6 0

a=Δv/Δt=(30.0-18.0)/8.0=12.0/8.0=1.5 m/s²

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Answer:

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We know that

Average acceleration,a= $\frac{v-u}{t}{t}$

Using the formula

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1 year ago

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

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Answer:

The value is $r = 5.077 \ m$

Explanation:

From the question we are told that

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Generally for the electron to be held up by the force gravity

Then

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2 years ago

When the boy crashes his bumper car into the girl's bumper car, the momentum from his car is transferred to hers. What evidence

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Answer:

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A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

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Initial velocity of the truck, $v_{Ti}=20\ m/s$

After the collision, velocity of the car, $v_{Cf}=18\ m/s$

Let $v$ is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

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$210000-21600=9000\ kg\times v$

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1 year ago

A particle of charge 2.3 ✕ 10−8 C experiences an upward force of magnitude 4.6 ✕ 10−6 N when it is placed in a particular point

Marysya12 [62]

<h2>Answer:</h2>

(a) +2 x 10² N/C (upwards)

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<h2>Explanation:</h2>

The force (F) acting on a particle of charge (Q) at a particular point is related to its electric field (E) by the following;

F = Q x E ----------------------(i)

This means that the force acting on the charged particle is the product of its charge and the electric field at that point.

<em>(a) Given</em>;

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<em>Substitute these values into equation (i) as follows;</em>

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=> E = 4.6 x 10⁻⁶ ÷ 2.3 x 10⁻⁸

=> E = 2 x 10² N/C

Since the value is positive, the electric field is directed upwards.

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<em>(b) If a charge of q = -1.1 x 10⁻⁸ is placed there, the force (F) acting is calculated as follows;</em>

<em>Substitute Q = q into equation (i) as follows;</em>

F = q x E

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F = -1.1 x 10⁻⁸ x 2 x 10²

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F = -2.2μN

Since the value of the force, F, is negative, it means it is directed downwards.

Therefore the force on the charge is -2.2μN

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2 years ago