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denis-greek [22]

3 years ago

9

Car 1 brakes to a stop on a dry road. Car 2 does the same thing, at the same speed, on a road where it has been raining. Explain

how the following will be different for car 2: a) braking distance, b) thinking distance and c) stopping distance.

1 answer:

Zepler [3.9K]3 years ago

8 0

Answer: car 1 is going how fast

Explanation: no need to answer without speed I won't know distance.

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A 75.5 kg diver drops from a diving board 10.0 m above the waters surface. Find the divers speed just before he strikes the wate

jenyasd209 [6]

Answer:

<em>v=14 m/s</em>

Explanation:

<u>Mechanical Energy </u>

The kinetic energy of a body (K) is the capacity of doing work due to its speed. It can be expressed as

$\displaystyle K=\frac{mv^2}{2}$

The potential energy (U) is the capacity of doing work due to its height respect to a certain reference level.

$U=mgh$

The mechanical energy is the sum of both

$\displaystyle E_m=\frac{mv^2}{2}+mgh$

The principle of conservation of mechanical energy states it must remain the same if no external force is acting on it. The diver drops from the diving board, which means its initial speed is zero (and so its initial kinetic energy). Thus, the mechanical energy at the jumping time is

$\displaystyle E_m=mgh=(75)(10)(9.8)=7350\ J$

When the diver is about to get into the water, his height reaches zero and the speed is at maximum. All the potential energy became kinetic energy, so

$\displaystyle \frac{mv^2}{2}=7350\ J$

Rearranging

$\displaystyle v^2=\frac{2(7350)}{75}=196$

$v=\sqrt{196}=14\ m/s$

The final speed of the diver is

$\boxed{v=14\ m/s}$

8 0

3 years ago

Which statement below BEST explains global warming?

vodka [1.7K]

The correct answer is A. an increase in Earth's average surface temperature.

While all of these are aspects of global warming, option A best summarizes it.

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2 years ago

1:a boy 2 a girl pulling a heavy crate at the same time w/10 unite of force each.What is the net force acting on the object

vovangra [49]

The maximum magnitude of the net force on the box is 20 N, which is only possible if the boy and the girl pull the box together in the same direction, horizontally and parallel to the ground.
The minimum magnitude of the net force on the box is 0 N, which will occur when the boy and the girl pull the box together in the parallel but opposite direction.
If either of them pulls at an angle from the horizontal, then the magnitude of the net force will be between 0 N and 20 N.

8 0

3 years ago

An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft

sergeinik [125]

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

<u>Given :</u>

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( $K_{A}$ = 0) to the point B at distance l = 0.500m where its kinetic energy is ( $K_{B}$ = 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

<u>Required :</u>

<em>(a) We are asked to find the electric potential VB </em>

<em>(b) We want to determine the magnitude and the direction of the electric field E. </em>

<u> Solution </u>

(a) We are given the values for VA, $K_{B}$ and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

$K_{A} +U_{A} =K_{B} +U_{B}$ .........................................(1)

Where $K_{A}$ = 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential. So, equation (1) will be in the form :

0+qVA= $K_{B}$ +qVB (Divide by q)

VA= $K_{B}$ /q + VB (solve for VB)

VB=VA- $K_{B}$ /q .......................................(2)

We get the relation between VB, VA and $K_{B}$ , now we can plug our values for VA, $K_{B}$ and q into equation (2) to get VB

VB=VA- $K_{B}$ /q

=30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

=80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

VA-VB = $\int\limits^1_0 {E} \, dl$ ...................................(a)

$=E\int\limits^1_0 {} \, dl$

VA-VB=E $l$ (solve for E)

E= VA-VB/ $l$ ..................................(3)

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

E= VA-VB/ $l$

=-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

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PolarNik [594]

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