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3 years ago

If you are in a car at a stop sign and then the driver

Physics

2 answers:

Cerrena [4.2K]3 years ago

7 0

The car is accelerating forward.

SVEN [57.7K]3 years ago

4 0

Answer:

it wont be because the hand brake would be on

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Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm

mihalych1998 [28]

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?

Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.

(a) What is the electric potential at point a due to q1 and q2?

(b) What is the electric potential at point b?

(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?

Answer:

a) the potential is zero at the center .

Explanation:

a) since the two equal-magnitude and oppositely charged particles are equidistant

b)(b) Electric potential at point b, v = Σ kQ/r

r = 5cm = 0.05m

k = 8.99*10^9 N·m²/C²

Q = -2 microcoulomb

v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m

v = -105 324 V

c)workdone = charge * potential

work = -6.00µC * -105324V

work = 0.632 J

6 0

3 years ago

Se lanza una piedra de 3.00 N verticalmente hacia arriba desde el suelo. Se observa que, cuando está 15.0 m sobre el suelo, viaj

BARSIC [14]

Answer:

(a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

Explanation:

Given that,

Weight of stone = 3.00 N

Height = 15 m

Speed = 25.0 m/s

(a). We need to calculate the speed at the moment of being thrown

Using work energy theorem

$W=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$-mg\times d=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

Put the value into the formula

$-9.8\times15=\dfrac{1}{2}\times(v_{2}^2-v_{1}^2)$

$-2\times9.8\times15=25^2-v_{1}^2$

$-v_{1}^2=-300-25^2$

$v_{1}=\sqrt{925}$

$v_{1}=30.41\ m/s$

(b). We need to calculate the maximum height

Using work energy theorem

$[tex]W=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2$

$mg\times d=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2$

Here, $\dfrac{1}{2}mv_{2}^2$ =0

$-(mg)\times d=\dfrac{1}{2}mv_{1}^2$

$d=\dfrac{v_{1}^2}{2g}$

Put the value into the formula

$d=\dfrac{30.41^2}{2\times9.8}$

$d=47.18\ m$

Hence, (a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

6 0

3 years ago

which would be better suited for agriculture, the soil of a tropical rain forest or that of a temperate deciduous forest

Romashka-Z-Leto [24]

I think it may be that of a temperate deciduous forest tho im not sure
thank u for letting me answer and god bless have a good life <3

6 0

3 years ago

A person pushes a 50kg box to the right with 100N of force at a constant velocity of 5m/s. Calculate the coefficient of friction

Anna11 [10]

The weight of the box is 50x9.8 = 490 N. The force of friction is 100N. F= μΝ so coefficient = 100/490 = 0.20

8 0

3 years ago

What are the potential out comes of force

AysviL [449]

Answer:

F in the definition of potential energy is the force exerted by the force field, e.g., gravity, spring force, etc. The potential energy U is equal to the work you must do against that force to move an object from the U=0 reference point to the position r.

Explanation:

5 0

2 years ago