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Dafna11 [192]
1 year ago
9

A 16 g ball at the end of a 1.4 m string is swung in a horizontal circle. It revolves once every 1.09 s. What is the magnitude o

f the string's tension?
Answer and I will give you brainiliest

​
Physics
1 answer:
olga nikolaevna [1]1 year ago
6 0

Answer:

6.56

Explanation:

frequency = 1/1.09 = 0.92 Hz

f = 1/(2L) × √T/m

0.92 = 1/2(1.4) × (√T/√16)

0.92 = 0.36 × (√T)/4

multiply all through by 4

3.68 = 1.44 × √T

√T = 3.68/1.44 = 2.56

T = 2.56² = 6.56

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A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
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Answer:

(a) A = 0.650 m

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(c) E = 17.1416 J

(d)  K = 11.8835 J

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Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

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A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.
mafiozo [28]

Answer:

a. f=1.22*10^{-15} N

b. f=53.6*10^{-17} N

Explanation:

The force existing between two charges is given as

f=\frac{kq_{1}q_{2}}{r^{2}}

where q= charge,

k=constant

r= distance between the two charges

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Also the charge of an electron is

-1.602*10^{-19}

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f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni

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for the -5.50nC the distance between them is 0.600m as can be seen in the diagram the magnitude of the force is

f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

The total net force on the electron is thus

f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\  f=1.22*10^{-15} N

b. at  distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

for the -5.50nC the distance between them is 0.4m as can be seen in the diagram the magnitude of the force is

f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.4^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.16}\\f_{23}=49.6*10^{-17}Ni

this this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

The total net force on the electron is thus

f=f_{21}+f_{23}\\ f=4.0*10^{-15}+49.6*10^{-17}\\  f=53.6*10^{-17} N

8 0
2 years ago
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