Answer:
conserved
Explanation:
During this process the energy is conserved
Answer:
The answer is not correct.
Explanation:
Stu's answer is not correct, the equation to use is known as the law of ohm. In which the voltage is defined as the product of the current by the resistance, then we will see this equation.
![V = I*R\\where:\\I = current [amp]\\R = resistance [ohm]\\V = voltage [volts]\\](https://tex.z-dn.net/?f=V%20%3D%20I%2AR%5C%5Cwhere%3A%5C%5CI%20%3D%20current%20%5Bamp%5D%5C%5CR%20%3D%20resistance%20%5Bohm%5D%5C%5CV%20%3D%20voltage%20%5Bvolts%5D%5C%5C)
In order to find resistance, this term is found multiplying the current on the right side of the equation, therefore the current will be divided on the left side of the equation.
![R=\frac{V}{I} \\replacing:\\R=\frac{4}{0.5} \\R=8[ohms]](https://tex.z-dn.net/?f=R%3D%5Cfrac%7BV%7D%7BI%7D%20%5C%5Creplacing%3A%5C%5CR%3D%5Cfrac%7B4%7D%7B0.5%7D%20%5C%5CR%3D8%5Bohms%5D)
That is the reason that the result found by Stu is not correct.
Answer:
a)693.821N/m
b)17.5g
Explanation:
We the Period T we can find the constant k,
That is
squaring on both sides,
where,
M=hanging mass, m = spring mass,
k =spring constant
T =time period
a) So for the equation we can compare, that is,
the hanging mass M is x here, so comparing the equation we know that
b) In order to find the mass of the spring we make similar process, so comparing,
<span>The angular momentum of a particle in orbit is
l = m v r
Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"
m_1 v_1 r_1 = m_2 v_2 r_2
Assuming that the mass did not change, conservation of angular momentum demands that
v_1 r_1 = v_2 r_2
or
v1 = v_2 (r_2/r_1)
Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have
v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s
Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
