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3 years ago

13

You walk forward 105 meters, in 15 seconds. What is your overall velocity? I need help immediately please

1 answer:

damaskus [11]3 years ago

8 0

Velocity = displacement/ time
105 meters/ 15 seconds = 7 m/s

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Spidermans nemesis electro delivers 4kj of electrical energy in half a second how much power does it draw from the mains?

Elodia [21]

Power delivered = (energy delivered) / (time to deliver the energy)

Power delivered = (4,000 J) / (0.5 sec)

Power delivered = 8,000 watts

I'm a little surprised to learn that Electro draws his power from the mains. This is VERY good news for Spiderman ! It means that Spiderman can always avoid tangling with Electro ... all he has to do is stay farther away from Electro than the length of Electro's extension cord.

But OK. Let's assume that Electro draws it all from the mains. Then inevitably, there must be some loss in Electro's conversion process, between the outlet and his fingertips (or wherever he shoots his bolts from).

The efficiency of Electro's internal process is

<em>(power he shoots out) / (power he draws from the mains) </em>.

So, if he delivers energy toward his target at the rate of 8,000 watts, he must draw power from the mains at the rate of

<em>(8,000 watts) / (his internal efficiency) . </em>

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4 years ago

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Consider five charged particles: A,B,C,D,and E.

Kisachek [45]

Answer:

A_negative

B_positive

C_positive

D_positive

E_negative

Explanation:

according to the law of electrostatic which is

like charge repels and unlike charge attract.

3 0

4 years ago

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Two tiny conducting spheres are identical and carry charges of -18.8 µC and +46.5 µC. They are separated by a distance of 2.47 c

sergiy2304 [10]

Answer:

$F=-12896N$ , attractive.

Explanation:

For calculating this force we use the Coulomb Law:

$F=\frac{kq_1q_1}{r^2}$

Where $k=9\times10^9Nm^2/C^{-2}$ is the Coulomb's constant, $q_1$ and $q_2$ the values of each charge and r the distance between them.

Since the Coulomb's constant as I wrote it is in S.I. we have to write all the magnitudes in that system of units, and substitute:

$F=\frac{(9\times10^9Nm^2/C^{-2})(-18.8\times10^{-6}C)(46.5\times10^{-6}C)}{(0.0247m)^2}=-12896N$

This force is attractive since both charges are of opposite sign.

3 0

3 years ago

The sum of the interior angles of a quadrilateral is _______. 180° 90° 270° 360°

nlexa [21]

Year9 formula: 180 (n-2)
Answer: 360

7 0

3 years ago

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A 1000-kg car is moving at 40.0 km/hr when the driver slams on the brakes and skids to a stop (with locked brakes) over a distan

liq [111]

Answer:

skid distance = 180 m

Explanation:

given,

mass of the car = 1000 Kg

speed of the car = 40 Km/hr

distance to stop = 20 m

if speed is now = 120 Km/he

distance to stop = ?

from given question we can state that,

Work done by the friction is equal to the change in kinetic energy of the car.

$W= \dfrac{1}{2}mv^2$ ............(1)

And we also know that work is directly proportional to displacement

W = F × x...........................(2)

from the equation (1) and (2)

x ∝ v²

now,

here velocity is increased from 40 km/h to 120 km/h means velocity is increase 3 times

so displacement will be equal to 3² or 9 times

hence, skid distance will be equal to 9 x 20 = 180 m

skid distance = 180 m

8 0

3 years ago