Answer:
B) 1.2 N, toward the center of the circle
Explanation:
The circumference of the circle is:
C = 2πr
C = 2π (0.70 m)
C = 4.40 m
So the velocity of the ball is:
v = C/t
v = 4.40 m / 0.60 s
v = 7.33 m/s
Sum of the forces in the radial direction:
∑F = ma
T = m v² / r
T = (0.015 kg) (7.33 m/s)² / (0.70 m)
T = 1.2 N
The tension force is 1.2 N towards the center of the circle.
<span>Answer:
Correct answer is
just add the two kinetic energies;
E = (1/2)mv^2 + (1/2)mv^2</span>
Solution:
According to the equations for 1-D kinematics. The only change to them is that instead one equation that describes general motion.
So we will have to use the equations twice: once for motion in the x direction and another time for the y direction.
v_f=v_o + at ……..(a)
[where v_f and v_o are final velocity and initial velocity, respectively]
Now ,
Initially, there was y velocity, however gravity began to act on the football, causing it to accelerate.
Applying this value in equation (a)
v_yf = at = -9.81 m/s^s * 1.75 = -17.165 m/s in the y direction
For calculating the magnitude of the equation we have to square root the given value
(16.6i - 17.165y)
\\
\left | V \right |=sqrt{16.6^{2}+17.165^{2}}\\ =
\sqrt{275.56+294.637225}\\=
\sqrt{570.197225}\\=
23.87[/tex]
Answer: Radiation
Explanation:
There are three ways in which the thermal transfer occurs:
1. By Conduction, when the transmission is by the <u>direct contact</u>.
2. By Convection, heat transfer <u>in fluids </u>(like water or the air, for example).
3. By Radiation, by the <u>electromagnetic waves</u> (they can travel through any medium and in <u>vacumm</u>)
So, in the outter space is vacuum, this means the energy cannot be transmitted by convection, nor conduction. It must be transmitted by electromagnetic waves that are able to travel with or without a medium, and this is called radiation.
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
