Answer:
The value is
Explanation:
From the question we are told that
The operating temperature is
The emissivity is
The power rating is
Generally the area is mathematically represented as
Where is the Stefan Boltzmann constant with value
So
<span>The velocity would be 54.2 m/s
We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.</span>
4. The Coyote has an initial position vector of .
4a. The Coyote has an initial velocity vector of . His position at time is given by the vector
where is the Coyote's acceleration vector at time . He experiences acceleration only in the downward direction because of gravity, and in particular where . Splitting up the position vector into components, we have with
The Coyote hits the ground when :
4b. Here we evaluate at the time found in (4a).
5. The shell has initial position vector , and we're told that after some time the bullet (now separated from the shell) has a position of .
5a. The vertical component of the shell's position vector is
We find the shell hits the ground at
5b. The horizontal component of the bullet's position vector is
where is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for :
Answer:
7.22 × 10²⁹ kg
Explanation:
For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.
So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km
The centripetal force F' = mRω² where R = radius of neutron star and ω = angular speed of neutron star
So, since F = F'
GMm/R² = mRω²
GM = R³ω²
M = R³ω²/G
Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²
Substituting the values of the variables into M, we have
M = R³ω²/G
M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²
M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²
M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²
M = 7217.66 × 10²⁶ kg
M = 7.21766 × 10²⁹ kg
M ≅ 7.22 × 10²⁹ kg