Answer:
im sure your already past this but it's E.
Explanation:
This is because in this case potential energy is linear to height, which means that the higher the more potential energy.
The magnitude of the electric field just outside the surface of the sphere is
![9 \times 10 {}^{4} n/c](https://tex.z-dn.net/?f=9%20%5Ctimes%2010%20%7B%7D%5E%7B4%7D%20n%2Fc)
given:
radius,r=0.01m
charge,Q=1.0×10-9
what is electric field?
Electric field is defined as the electric force per unit charge.It is surrounded by electrically charged particles and exerts force on all other charge particles.
using formula,
![E = KQ/r {}^{2}](https://tex.z-dn.net/?f=E%20%3D%20KQ%2Fr%20%7B%7D%5E%7B2%7D%20)
for just outside r=radius of sphere
![E = \frac{9 \times 10 {}^{9} \times 10 {}^{ - 9} }{(0.01){}^{2} }](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B9%20%5Ctimes%2010%20%7B%7D%5E%7B9%7D%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%209%7D%20%20%7D%7B%280.01%29%7B%7D%5E%7B2%7D%20%7D%20)
![= \frac{9 \times 10 {}^{9} \times 10 {}^{ - 9} }{0.0001}](https://tex.z-dn.net/?f=%20%3D%20%20%5Cfrac%7B9%20%5Ctimes%2010%20%7B%7D%5E%7B9%7D%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%209%7D%20%20%7D%7B0.0001%7D%20)
![= 9 \times 10 {}^{4} n/c](https://tex.z-dn.net/?f=%20%3D%209%20%5Ctimes%2010%20%7B%7D%5E%7B4%7D%20n%2Fc)
Thus,the electric field just outside the surface of the sphere is
![= 9 \times 10 {}^{4} n/c](https://tex.z-dn.net/?f=%20%3D%209%20%5Ctimes%2010%20%7B%7D%5E%7B4%7D%20n%2Fc)
learn more about electric field from here: brainly.com/question/28203588
#SPJ4
Answer:
1.05 J.
Explanation:
Kinetic Energy: This is the energy possessed by a body due to its motion. The S.I unit of kinetic energy is Joules (J). The formula of kinetic energy is given as
Ek = 1/2mv²................. Equation 1
Where Ek = kinetic energy, m = mass of the uniform rod, v = liner velocity of the rod.
But,
v = αr .......................... Equation 2
Where α = angular velocity of the rod, r = radius of the circle.
Given: α = 3.6 red/s, r = 120/2 = 60 cm = 0.6 m.
Substitute into equation 2
v = 3.6(0.6)
v = 2.16 m/s.
Also given: m = 450 g = 0.45 kg.
Substitute into equation 1
Ek = 1/2(0.45)(2.16²)
Ek = 1.05 J.
Answer: A. The total displacement divided by the time and C. The slope of the ant's displacement vs. time graph.
Explanation:
Hi! The question seems incomplete, but I found the options on the internt:
A. The total displacement divided by the time.
B. The slope of the ant's acceleration vs. time graph.
C. The slope of the ant's displacement vs. time graph.
D. The average acceleration divided by the time.
Now, since we know the ant is travelling at a constant speed, its average velocity
will be expressed by the following equation:
![V=\frac{d}{t}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7Bd%7D%7Bt%7D)
Where:
is the ant's total displacement
is the time it took to the ant to travel to the kitchen
Hence one of the correct options is: A. The total displacement divided by the time
On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is:
C. The slope of the ant's displacement vs. time graph.
Answer:
F = 1.58*10^{11} N
Explanation:
given data:
length of steel beam = 27 m
cross sectional area of rail = 35 cm
Degree celcius
change in length of steel beam is given as
![\Delta L = L_O \alpha \Delta T](https://tex.z-dn.net/?f=%5CDelta%20L%20%3D%20L_O%20%5Calpha%20%5CDelta%20T)
![= 20*1.1*10^{-5}*39](https://tex.z-dn.net/?f=%3D%2020%2A1.1%2A10%5E%7B-5%7D%2A39)
m
Young's modulus is
![Y = \frac{FL}{A\Delta L}](https://tex.z-dn.net/?f=Y%20%3D%20%5Cfrac%7BFL%7D%7BA%5CDelta%20L%7D)
![F = \frac{ YA\Delta L}{L}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%20YA%5CDelta%20L%7D%7BL%7D)
![= \frac{2.0*10^{11}*25*10^{-4}8.58*10^{-3}}{27}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B2.0%2A10%5E%7B11%7D%2A25%2A10%5E%7B-4%7D8.58%2A10%5E%7B-3%7D%7D%7B27%7D)
F = 1.58*10^{11} N