Answer:
22.36 rad
Explanation:
Applying,
ω = θ/t.............. Equation 1
Where ω = angular velocity, θ = angular displacement of the baseball, t = time
make θ the subject of the equation
θ = ωt............... Equation 2
From the question,
Given: ω = 350 rev/min = 350(0.10472) = 36.652 rad/s, t = 0.61 s
Substitute these values into equation 1
θ = 0.61(36.652)
θ = 22.36 rad
Hence the angular displacement of the baseball is 22.36 rad
Below are the 5 main indicators of chemical change.
Chemical change indicators:<span>
Color change
</span>Temperature change
Precipitate formation<span>
Odor
Bubble formation
I hope this helps!</span>
Answer:
![125\sqrt[4]{8}](https://tex.z-dn.net/?f=125%5Csqrt%5B4%5D%7B8%7D)
Explanation:
A number of the form

can be re-written in the radical form as follows:
![\sqrt[n]{a^m}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D)
In this problem, we have:
a = 1,250
m = 3
n = 4
So, if we apply the formula, we get
![1,250^{\frac{3}{4}}=\sqrt[4]{(1,250)^3}](https://tex.z-dn.net/?f=1%2C250%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%3D%5Csqrt%5B4%5D%7B%281%2C250%29%5E3%7D)
Then, we can rewrite 1250 as

So we can rewrite the expression as
![=\sqrt[4]{(2\cdot 5^4)^3}=5^3 \sqrt[4]{2^3}=125\sqrt[4]{8}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B4%5D%7B%282%5Ccdot%205%5E4%29%5E3%7D%3D5%5E3%20%5Csqrt%5B4%5D%7B2%5E3%7D%3D125%5Csqrt%5B4%5D%7B8%7D)
C......................................
Answer:


Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/r²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
r: distance from charge q to point P in meters (m)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Equivalences
1nC= 10⁻⁹ C
1cm= 10⁻² m
Graphic attached
The attached graph shows the field due to the charges:
Ep₁: Total field at point P due to charge q₁. As the charge is positive ,the field leaves the charge.
Ep₂: Total field at point P due to charge q₂. As the charge is negative, the field enters the charge.
Known data
q₁ = 63 nC = 63×10⁻⁹ C
q₂ = -47 nC = -47×10⁻⁹ C
k = 8.99*10⁹ N×m²/C²
d₁ = 1.4cm = 1.4×10⁻² m
d₂ = 3.4cm = 3.4×10⁻² m
Calculation of r and β


Problem development
Ep: Total field at point P due to charges q₁ and q₂.

Ep₁ₓ = 0



Calculation of the electric field components at point P

