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3 years ago

PLEASE HELP!!! SCIENCE QUESTION!

Physics

2 answers:

Tatiana [17]3 years ago

5 0

C. has to be right, also its the only one that makes sense.
Hope this helps!

Cerrena [4.2K]3 years ago

4 0

Answer:

C. Electronic charges have electric fields surrounding them that allow them to exert forces on other objects without touching them.

Explanation:

All the electric charges have electric field in the region surrounding them

This electric field is given as

$E = \frac{kq}{r^2}$

above formula is the electric field due to a point charge in the surrounding region of it.

Now due to this electric field when other charges comes into this region they will experience electric force on them which is given as

$F = qE$

due to above reason two charges either attract or repel each other

so correct answer will be

C. Electronic charges have electric fields surrounding them that allow them to exert forces on other objects without touching them.

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J-s. If your 1400-kg car is parked in an 8.54-m-long garage, what is the uncertainty in its velocity? cm/s the tolerance is +/-2

IgorC [24]

Answer:

$\Delta v = 4.41 \times 10^{-37} cm/s$

Explanation:

As per Heisenberg's uncertainty principle we know that

$\Delta P \times \Delta x = \frac{h}{4\pi}$

so here we have

$\Delta P = m\Delta v$

$\Delta x = 8.54 m$

now from above equation we have

$m\Delta v \times (8.54) = \frac{h}{4\pi}$

$1400(\Delta v) \times (8.54) = \frac{6.626 \times 10^{-34}}{4\pi}$

$\Delta v = 4.41 \times 10^{-39} m/s$

$\Delta v = 4.41 \times 10^{-37} cm/s$

8 0

3 years ago

Calculate the acceleration of a 7,000 kg, single-engine airplane just before takeoff when the thrust of its

Arada [10]

The engine is 10,000 N I think. Hopefully this helps!!

3 0

3 years ago

A point charge q1 = 4.50 nC is located on the x-axis at x = 1.95 m , and a second point charge q2 = -6.80 nC is on the y-axis at

Vinvika [58]

Answer:

Explanation:

One charge is situated at x = 1.95 m . Second charge is situated at y = 1.00 m

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Applying Gauss's theorem

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3 0

3 years ago

A student pulls down with a force of 40 N on a rope that winds around a pulley of radius 5 cm. find the torque?

Anna71 [15]

Answer:

Explanation:

Given that a student pulls down with a force of 40 N on a rope that winds around a pulley of radius 5 cm. find the torque?

Solution

Using the formula

Torque = Force × radius

I.e

Torque = F r

Where

F = 40N

r = 5 cm = 5/100 = 0.05m

Substitute all the parameters into the formula

Torque = 40 × 0.05

Torque = 2 N

Therefore, the torque on this scenario is 2 N

7 0

3 years ago

Help me frfr I don’t understand

Art [367]

The points are

(1,10)

(6,0)

$\boxed{\sf slope(m)=\dfrac{y_2-y_1}{x_2-x_1}}$

$\\ \sf\longmapsto m=\dfrac{0-10}{6-1}$

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6 0

3 years ago