All categories

2 years ago

Protons Neutrons Electrons Location Charge Size

Physics

1 answer:

igor_vitrenko [27]2 years ago

6 0

Protons are positive, and neutrons are negative, electrons are neutral. I’m not sure about the rest but I hope that helps for now

You might be interested in

If a ball is thrown into the air with a velocity of 36 ft/s, its height (in feet) after t seconds is given by y = 36t − 16t2. fi

Juliette [100K]

Height (y) = 36t - 16t^2, where t = time in seconds (s).
Our height (y) after 1s = 36(1) - 16(1)^2
y = 36 - 16 = 20 ft
So it reached a height of 20 ft during that 1 second, which means that at that 1 second it had a velocity of 20ft/s, since v = d(distance)/t = 20ft/1s

6 0

3 years ago

Which graph accurately shows the relationship between kinetic energy and speed as speed increases?

mixer [17]

Answer:

Explanation:

kinetic energy (KE) is the energy possessed by moving bodies. It can be expressed as:

KE = $\frac{1}{2}$ m $v^{2}$

Where: m is the mass of the object, and v its speed.

For example, a stone of mass 2kg was thrown and moves with a speed of 3 m/s. Determine the kinetic energy of the stone.

Thus,

KE = $\frac{1}{2}$ x 2 x $3^{2}$

= 9

KE = 9.0 Joules

Assume that the speed of the stone was 4 m/s, then its KE would be:

KE = $\frac{1}{2}$ x 2 x $4^{2}$

= 16

KE = 16.0 Joules

Therefore, it can be observed that as speed increases, the kinetic energy increases. Thus option B is appropriate.

3 0

3 years ago

The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°c _____.

Fiesta28 [93]

Answer:

4.2 J

Explanation:

Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K

From specific heat capacity,

Q = cmΔt.............................. Equation 1

Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.

Given: m = 1 g = 0.001 kg, Δt = 1 °C

Constant : c = 4200 J/kg.°C

Substitute into equation 1

Q = 0.001×4200(1)

Q = 4.2 J.

Hence the energy absorbed or lost = 4.2 J

6 0

3 years ago

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

wariber [46]

Answer:

For left = 0 N/C

For right = 0 N/C

At middle = $-7.6836 * 10^{-11} \vec{i}$ N/C

Explanation:

Given data :-

б = $6.8 * 10^{-22}$ C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

$E = \frac{\sigma }{2\varepsilon }$

1) To the left of the plate

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

2) To the right of them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

3) Between them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(-\vec{i})$ = $(\frac{\sigma }{\varepsilon })(-\vec{i})$ = $-\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} } \vec{i}$ = $-7.6836 * 10^{-11} \vec{i}$ N/C

5 0

3 years ago

1.<br>Force that pulls toward the center of mass.<br><br>

kondaur [170]

Gravity if I’m not mistaken

7 0

3 years ago

Read 2 more answers