If the sun were to collapse into a black hole, what would the radius be?
Answer:
If the Sun, with its mass of 1 MSun, were to become a black hole the Schwarzschild radius would be about 3 kilometers; thus, the entire black hole would be about one-third the size of a neutron star of that same mass
Answer:
It would be 1/3rd size of a neutron star
Explanation:
the Schwarzschild radius would be about 3 kilometers; thus, the entire black hole would be about one-third the size of a neutron star of that same mass.
Sam is moving house and is carrying a 300N box of books up a flight of steps 5m high, it takes her 30 seconds. Gary follows her carrying a bag of clothes doing 1000 J of work; it only takes him 25 seconds. Who provides the biggest power? Show your working.
Answer: Sam provides the biggest power
Explanation: Power is defined, in Physics, as the rate of work done. Work is energy transferred to an object due to the force causing the displacement of the object.
So, to determine Power:
For Sam:
P = [tex]\frac{work}{time}[/tex]
[tex]P=\frac{F.d}{t}[/tex]
[tex]P=\frac{300.5}{30}[/tex]
P = 50 Watts
The unit for Power is [P] = J/s = Watt
For Gary:
P = [tex]\frac{work}{time}[/tex]
[tex]P=\frac{1000}{25}[/tex]
P = 40 Watts
Comparing power of Sam and Gary, we can conclude that Sam provides the biggest power of 50 Watts.
is there any possibility to make 100% efficient system
Answer:
yes
Explanation:
its awesome
2. Two projectiles thrown from the same point at angles 0,=60° and 02=30° with the horizontal,
attain the same height. What is the ratio of their initial velocities (Voi:Voz)?
Answer:
The answer is below
Explanation:
The maximum height (h) of a projectile with an initial velocity of u, acceleration due to gravity g and at an angle θ with the horizontal is given as:
[tex]h=\frac{u^2sin^2\theta}{2g}[/tex]
Given that the two projectile has the same height.
[tex]For\ the\ first\ projectile\ with\ an\ angle\ \60^0 \ with\ the\ horizontal\ and initial\ velocity\ u_1:\\\\h=\frac{u_1^2sin^260}{2g} =\frac{0.75u_1^2}{2g} \\\\For\ the\ second\ projectile\ with\ an\ angle\ \30 \ with\ the\ horizontal\ and initial\ velocity\ u_2:\\\\h=\frac{u_2^2sin^230}{2g} =\frac{0.25u_2^2}{2g}\\\\\frac{0.25u_2^2}{2g}=\frac{0.75u_1^2}{2g}\\\\0.25u_2^2=0.75u_1^2\\\\\frac{u_1^2}{u_2^2} =\frac{0.25}{0.75} \\\\\frac{u_1^2}{u_2^2}=\frac{1}{3} \\[/tex][tex]\\\frac{u_1}{u_2}=\sqrt\frac{1}{3}[/tex]
77. Two blocks, with masses m1 = 400 g and m2 = 600 g, are connected by a string and lie on a frictionless tabletop. A force F = 3.5 N is applied to block m2. Find the acceleration of each object.
Answer:
See the answers below. And the free body diagrams attached.
Explanation:
To solve this problem, we must build a free body diagram, where we show the forces acting on a body, first for the m2 body, where the force of 3.5 [N] acts to the right. In such a way that a reaction force is presented in the opposite direction due to the string tied between both bodies, we will call this force T1.
Now using Newton's second law, which is defined as the sum of forces equal to the product of mass by acceleration, we can determine an equation as a function of T1.
∑F = m*a
[tex]3.5 -T_{1}=0.6*a_{2}[/tex]
We have an equation with two unknowns, in such a way we must perform a second free body diagram for the body with mass m1, to acquire the additional equation.
∑F =m₁*a₁
[tex]T_{1}=0.4*a_{1}[/tex]
By kinematics we know that if the string tied to Body 2 moves a distance, the part of the string at the end tied to body 1 will move the same distance. This same analysis is valid for velocities and accelerations.
This is the accelerations a1 and a2 are equal. Now equalizing both equations.
[tex]3.5-0.4*a_{1}=0.6*a_{2}\\but\\a_{1}=a_{2}\\3.5-0.4*a_{1}=0.6*a_{1}\\a_{1}=3.5[m/s^{2} ][/tex]
The acceleration of each block connected by the given string is 3.5 m/s².
The given parameters;
mass of the first block, m1 = 400 g = 0.4 kgmass of the second block, m2 = 600 g = 0.6 kgapplied force, F = 3.5 NThe net force on the blocks is calculated as follows;
[tex]\Sigma F = ma\\\\F - T = m_2a\\\\F - m_1(a -g)= m_2a\\\\g = 0 \ , \ in \ horizontal \ direction\\\\F - m_1a = m_2 a\\\\F = m_2 a + m_1 a\\\\F = a(m_1 + m_2)\\\\a = \frac{F}{m_1 + m_2} \\\\a = \frac{3.5 }{0.4 + 0.6} \\\\a = 3.5 \ m/s^2[/tex]
Thus, the acceleration of each block connected by the given string is 3.5 m/s².
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6. What charge is in the nucleus?
a. negative charge
b. positive charge
C. no charge
d. positive and negative charges
