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Blizzard [7]
3 years ago
14

A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the obje

ct and (b) the magnitude of the resultant force.
Physics
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

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The velocities of light in air and glass are 3.0 x 10^8ms and 2.0×10^8ms respectively. If the angle of refraction is 30°, the si
JulijaS [17]

Answer:

0.75

Explanation:

refractive \: index \:  =  \frac{3.0 \times  {10}^{2} }{2.0 \times  {10}^{2} }

= 1.5

refractive \: index =  \frac{ \sin(angle \: of \: incidence) }{ \sin(angle \: of \: refraction) }

1.5 =  \ \frac{ \sin(i) }{ \sin(30) }

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5 0
3 years ago
The gas pressure inside a container decreases when
avanturin [10]

Answer:

When the volume increases or when the temperature decreases

Explanation:

The ideal gas equation states that:

pV= nRT

where

p is the gas pressure

V is the volume

n is the number of moles of gas

R is the gas constant

T is the gas temperature

Assuming that we have a fixed amount of gas, so n is constant, we can rewrite the equation as

\frac{pV}{T}=const.

which means the following:

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3 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
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He must travel 35 mph
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