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2 years ago

At a race car driving event, a staff member notices that the skid marks left by the race car are

1 answer:

5 0

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).

We can calculate the initial velocity of the car (u) using the following kinematic equation.

$v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s$

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

Learn more: brainly.com/question/14851168

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What increases the work output of a machine

Ilia_Sergeevich [38]

There are only two things you can do to increase the work output of a machine:

1). Increase the work INput to the machine.

2). Make the machine more efficient ... do things like lubricating it better to eliminate some internal friction.

4 0

3 years ago

A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t

kodGreya [7K]

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

$W=Fd cos \theta$ (1)

where

F is the magnitude of the force

d is the distance covered by the object

$\theta$ is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

$\theta=0$

And the distance covered is equal to the circumference of the circle, which is:

$d=2\pi r=2\pi (2.5 m)=15.7 m$

where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

$W=(3.0)(15.7)(cos 0)=47.1 J$

Learn more about work:

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2 years ago

Some superconductors are capable of carrying a very large quantity of current. If the measured current is 1.00 ´ 105 A, how many

Zielflug [23.3K]

Answer:

The $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

Explanation:

Given :

Current $I = 1 \times 10^{5}$ A

Charge of electron $e = 1.6 \times 10^{-19}$ C

Time $t = 1$ sec

From the formula of current,

Current is the number of charges flowing per unit time.

$I = \frac{ne}{t}$

Where $n =$ number of charges means in our case number of electrons

$n = \frac{It}{e}$

$n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }$

$n = 6.25 \times 10^{23}$

Therefore, $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

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3 years ago

You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle

nikitadnepr [17]

Answer:

a. cosθ b. E.A

Explanation:

a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for 90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function

b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A