Answer:
(a). The required is 1871.2 J.
(b). The speed the lead bullet is 395 m/s.
(c). The 20% of energy must have gone into collision between steel plate and bullet.
Explanation:
Given that,
Mass of bullet = 0.030 kg
Temperature = 20°C
(a). We need to heat required to increase the temperature of the lead bullet and melt it
Using formula of heat
Where, m = mass of lead bullet
S = specific heat
L = latent heat
T = Temperature
Put the value into the formula
The required is 1871.2 J.
(b). Assume that 80% of the bullet’s kinetic energy goes into increasing its temperature and then melting it
We need to calculate the energy
We need to calculate the speed the lead bullet
Using formula of speed
The speed the lead bullet is 395 m/s.
(c). The 20% of energy must have gone into collision between steel plate and bullet.
Hence, This is the required solution.
<span>covalent bond I think </span>
Explanation:
Ohm's law describes the relationship between voltage, current, and resistance.
V = IR
where V is voltage, I is current, and R is resistance.
A. At the original voltage:
V₁ = I₁ R₁
When the voltage is doubled and resistance stays the same:
2V₁ = I₁' R₁
Dividing the two equations:
2V₁ / V₁ = (I₁' / I₁) (R₁ / R₁)
2 = I₁' / I₁
So the new current is double the original current.
B. At the original voltage and resistance:
V₂ = I₂ R₂
When both the voltage and resistance are increased by a factor of 2:
2V₂ = I₂' (2R₂)
Dividing the two equations:
(2V₂ / V₂) = (I₂' / I₂) (2R₂ / R₂)
2 = (I₂' / I₂) (2)
1 = I₂' / I₂
So the new current is the same as the original current.
Answer: It would increase.
Explanation:
The equation for determining the force of the gravitational pull between any two objects is:
Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.
Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.
Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.
