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const2013 [10]
2 years ago
12

If a person runs 720m/minute for 15 minutes, how far will the person run?

Physics
1 answer:
julsineya [31]2 years ago
3 0

Answer:

10800

Explanation:

if someone runs 720 miles every minutes they will run 10,800 miles in 15 minutes

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What is the net force on a car with a mass of 1000 kg if its<br> acceleration is 35 m/s^2?
VashaNatasha [74]

Answer:

3000N

Explanation:

divided to get answer

the force needed to accelerate the 1000kg car by 3m/s2 is 3000N

7 0
2 years ago
For Valentine’s Day, Sally received a helium-filled balloon at a party. On returning home she accidentally left the balloon in t
Softa [21]

Answer:

If the temperature of the air in the balloon is less than the temperature of the air surrounding the balloon then the balloon will appear slightly deflated because of the difference in temperature.

As the temperature of the air in the balloon reaches the surrounding air temperature, then the balloon will appear to be fully inflated because the temperature of the air in the balloon is the same as the surrounding air temperature.

8 0
3 years ago
A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.
marishachu [46]

Answer:

a) the magnitude of the force is

F= Q(\frac{kqs}{r^3}) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b)  the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

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

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = \frac{kp}{r^{3} } , where p = q × s

E(r) = \frac{kqs}{r^{3} } where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q(\frac{kqs}{r^3})

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall  F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

8 0
3 years ago
Which statement is true about effective nuclear charge?a. Effective nuclear charge decreases as we move to the right across a ro
MissTica

Answer:

Option b. Effective nuclear charge increases as we move to the right across a row in the periodic table

Explanation:

The <em>effective nuclear charge </em>is a measure of how strong the protons in the nucleus of an atom attract the outermost electrons of such atom.

The <em>effective nuclear charge</em> is the net positive charge experienced by valence electrons and is calculated (as an approximation) by the equation: Zeff = Z – S, where Z is the atomic number and S is the number of shielding electrons.

The shielding electrons are those electrons in between the interesting electrons and the nucleus of the atom.

Since the shielding electrons are closer to the nucleus, they repel the outermost electrons and so cancel some of the attraction exerted by the positive charge of the nucleus, meaning that the outermost electrons feel less the efect of attraction of the protons. That is why in the equation of Zeff, the shielding electrons (S) subtract the total from the atomic number Z.

The <em>effective nuclear charge</em>, then, is responsible for some properties and trends in the periodic table. Here, you can see how this explains the trend of the atomic radius (size of the atom) accross a row in the periodic table.

  • As the<em> effective nuclear charge</em> is larger, in a same row of the periodic table, the shielding effect is lower, the outermost electrons are more strongly attracted by the nucleus, and the size of the atoms decrease. That is why as we move to the right in the periodic table, the size of the atoms decrease.

3 0
3 years ago
PLEASE HELP! **
Eduardwww [97]
It moves to 56 km per hours
8 0
3 years ago
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