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4 years ago

If an atom has 17 protons, 14 neutrons, and 21 electrons, what is the atom's electrical charge?

Physics

1 answer:

nlexa [21]4 years ago

8 0

Answer:

the electrical charge would be -4

Explanation:

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In a double-slit experiment, it is observed that the distance between adjacent maxima on a remote screen is 1.0 cm. What happens

kobusy [5.1K]

Answer:

C) It increases to 2.0 cm

Explanation:

In a double-slit diffraction experiment, the distance on the screen between two adjacent maxima is given by

$\Delta y = \frac{\lambda D}{d}$

where

$\lambda$ is the wavelength of the wave

D is the distance of the screen from the slits

d is the separation between the slits

In this problem, the initial distance between adjacent maxima is 1.0 cm. Later, the slit separation is cut in a half, which means that the new slit separation is

$d'=\frac{d}{2}$

Substituting into the equation, we find that the new separation between the maxima is

$\Delta y' = \frac{\lambda D}{d/2}=2(\frac{\lambda D}{d})=2\Delta y$

So, the distance increases by a factor 2: therefore, the new separation between the maxima will be 2.0 cm.

5 0

3 years ago

Can some one please help me

natta225 [31]

Answer:

its "up, then down, then up"

Explanation:

8 0

3 years ago

Two identical cars are driving along the motorway. Car A is travelling at 50 km/h and Car B is travelling at 80km/h. Which car h

maxonik [38]

Answer:

i think it is car B because the greater the speed, the greater the drag force acting on the vehicle

Explanation:

6 0

3 years ago

Calculating work for different springs Calculate the work required to stretch the following springs 0.5 m from their equilibrium

Rina8888 [55]

Answer:

Part a)

U = 31.25 J

Part b)

U = 312.5 J

Explanation:

Part A)

A spring that requires a force of 50 N to be stretched 0.2 m from its equilibrium position.

So here we have

$F = kx$

$50 = k(0.2)$

k = 250 N/m

now the energy stored in the spring is given by

$U = \frac{1}{2}kx^2$

$U = \frac{1}{2}(250)(0.5)^2$

$U = 31.25 J$

Part B)

A spring that requires 50 J of work to be stretched 0.2 m from its equilibrium position.

So here we know the formula of spring energy as

$U = \frac{1}{2}kx^2$

$50 = \frac{1}{2}k(0.2)^2$

$k = 2500 N/m$

now by the formula of energy stored in spring

$U = \frac{1}{2}kx^2$

$U = \frac{1}{2}(2500)(0.5)^2$

$U = 312.5 J$

6 0

3 years ago

Plzz answer correctly

Wittaler [7]

Answer:

Explanation:

5 0

3 years ago