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3 years ago

Which statement about protons is true?

2 answers:

8 0

Answer to your question is A. Because you cannot have HALF of a Proton...
So of course you will always have a whole number...

White raven [17]3 years ago

4 0

GIVE vo0Mac0ov Brainliest ... He's smart

You might be interested in

To find the number of neutrons in an atom, you would subtract

Luda [366]

Atomic number is equal to the number of protons and electrons

Atomic mass - protons = neutrons

protons + neutrons = atomic mass

I hope this helps

3 0

3 years ago

A particle of mass 2kg resting on a smooth table attached to a fixed point on the table by a rope 1.0m making 300revolution per

zepelin [54]

Answer is: 1973.17N aprox.
step by step in the pic below

7 0

3 years ago

A kettle is rated at 1 kW, 220 V. Calculate the working resistance of the kettle.

Anna [14]

Explanation:

Power of electric kettle, P = 1 kW

Voltage, V = 220 V

(a) Electric power is given by the formula as follows :

$P=\dfrac{V^2}{R}$

R is resistance

$R=\dfrac{V^2}{P}\\\\R=\dfrac{(220)^2}{10^3}\\\\R=48.4\ \Omega$

(b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.

Energy supplied is given by :

$E=P\times t$

P is power, $P=\dfrac{V^2}{R}$

$E=\dfrac{V^2}{R}t\\\\E=\dfrac{(220)^2}{48.4}\times 180\\\\E=180000\ J\\\\E=180\ kJ$

5 0

3 years ago

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (5*10^{-3})(300)^2$

$KE = 225J$

On the other hand the required Energy to heat up t melting point is

$Q_1 = mC_p \Delta T$

$Q_2 = L_f m$

Where,

m = Mass

$C_p =$ Specific Heat

$\Delta T =$ Change at temperature

$L_f =$ Latent heat of fussion

Heat required to heat up to melting point,

$Q = Q_1+Q_2$

$Q = mC_p \Delta T+L_f m$

$Q = 5*0.128*(327-20) + 5*24.7$

$Q = 310J$

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0

3 years ago

A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a

Zolol [24]

Considering the unknown resistence as R and using the Ohm's First Law, we have:

$i= \frac{V}{R_{eq}} \\ 0.5= \frac{12}{R+10} \\ R+10=24 \\ R=14-Ohm$

The equivalent resistence is given by the resistor series with the lamp resistence.

$R_{eq}=R+10 \\ R_{eq}=14+10 \\ \boxed {R_{eq}=24-Ohm}$

If you notice any mistake in my english, please let me know, because i am not native.

6 0

2 years ago