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3 years ago

Could a nucleus that has one proton but no neutrons exist?

1 answer:

8 0

Yes, but there is only 1 atom like that and is is hydrogen. Hydrogen is the only element that could have a nucleus with one proton and no neutrons exist.

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The resistance created by waves on a 120-m-long ship is tested in a channel using a model that is 4 m long Y Part A If the ship

DanielleElmas [232]

Answer:

$V_m = 12.78 km/hr$

Explanation:

given,

length of the ship = 120 m

length of model of the ship = 4 m

Speed at which the ship travels = 70 km/h

speed of model = ?

by using froude's law

$F_r = \dfrac{V}{\sqrt{L g}}$

for dynamic similarities

$(\dfrac{V}{\sqrt{L g}})_P = (\dfrac{V}{\sqrt{L g}})_{model}$

$(\dfrac{V_p}{\sqrt{L_p}}) = (\dfrac{V_m}{\sqrt{L_m}})$

$(\dfrac{70}{\sqrt{120}}) = (\dfrac{V_m}{\sqrt{4}})$

$V_m = 12.78 km/hr$

hence, the velocity of model will be 12.78 km/h

6 0

3 years ago

Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determin

tia_tia [17]

Answer with Explanation:

We are given that

A=3i-3j m

B=i-4 j m

C=-2i+5j m

a. $D=A+B+C$

$D=3i-3j+i-4j-2i+5j$

$D=2i-2j$

Compare with the vector r=xi+yj

We get x=2 and y=-2

Magnitude= $\mid D\mid=\sqrt{x^2+y^2}=\sqrt{(2)^2+(-2)^2}=2\sqrt 2$ units

By using the formula $\mid r\mid=\sqrt{x^2+y^2}$

Direction: $\theta=tan^{-1}\frac{y}{x}$

By using the formula

Direction of D: $\theta=tan^{-1}(\frac{-2}{2})=tan^{-1}(-1)=tan^{-1}(-tan45^{\circ})=-45^{\circ}$

b.E=-A-B+C

$E=-3i+3j-i+4j-2i+5j$

$E=-6i+12j$

$\mid E\mid=\sqrt{(-6)^2+(12)^2}=13.4$ units

Direction of E= $\theta=tan^{-1}(\frac{12}{-6}=tan^{-1}(-2)=-63.4^{\circ}$

4 0

3 years ago

Một cái đĩa có bán kính R quay quanh trục vuông góc và đi qua tâm đĩa theo phương trình ϕ = at3 + bt2 + c;Tìm gia tốc tiếp tuyến

Rama09 [41]

Answer:

Gnlhjsfyjjgisiidisc. Gihiingiifciucuc

5 0

2 years ago

Read 2 more answers

What does it mean for a chemical equation to be balanced?

kolezko [41]

Answer:

<h2>The number of each type of atom is the same for both the reactants and products.</h2>

I hope this helps

3 0

3 years ago

The gauge pressure in the tires of your car is 210 kPa (30.5 psi) when the temperature is 25°C (77 °F). Several days later it is

DaniilM [7]

Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:

P/T = const.

P = pressure, T = temperature, the quotient of P/T must stay constant.

Initial P and T values:

P = 210kPa + 101.325kPa

P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)

T = 25°C = 298.15K

Final P and T values:

P = ?, T = 0°C = 273.15K

Set the initial and final P/T values equal to each other and solve for the final P:

311.325/298.15 = P/273.15

P = 285.220kPa

Subtract 101.325kPa to find the final gauge pressure:

285.220kPa - 101.325kPa = 183.895271kPa

The final gauge pressure is 184kPa or 26.7psi.

8 0

3 years ago