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horrorfan [7]
3 years ago
5

Heather and Matthew walk with an average velocity of 0.98 m/s eastward. If it takes them 34 min to walk to the store, what is th

eir displacement?
Physics
1 answer:
TEA [102]3 years ago
3 0

Answer:

D= 1999.2 m

Explanation:

Given that

Average velocity ,v= 0.98 m/s

time ,t= 34 min

We know that

1 min  = 60 s

That is why

t= 34 x 60 =2040 s

We know that

Displacement = Average velocity x time

D= v t

Now by putting the values in the above equation

D= 0.98 x 2040 m

D= 1999.2 m (eastward)

The direction of the displacement will be towards eastward.

That is why the displacement will be 1999.2 m or we can say that 1.9992 km.

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A charge of 8.5 × 10–6 C is in an electric field that has a strength of 3.2 × 105 N/C. What is the electric force acting on the
Sonja [21]

2.72 N

Explanation:

Step 1:

From the basic formula in electrostatics

F = E * q

where F = Force due to charges

           E = Electric field strength

           q = Charge

Step 2:

From the given question

q= 8.5*10^{-6} C

E = 3.2 * 10^{5} N/C

F = 8.5 * 10^{-6} * 3.2 * 10^{5} = 2.72 N

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4 years ago
A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the
Julli [10]

Answer:

<h2>a) Q = 0.759µC</h2><h2>b) E = 39.5µJ</h2>

Explanation:

a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV

C = capacitance of the capacitor (in Farads )

V = voltage (in volts) = 100V

C = ∈A/d

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A = cross sectional area = 600 cm²

d= distance between the plates = 0.7cm

C = 8.85 × 10^-12 * 600/0.7

C = 7.59*10^-9Farads

Q = 7.59*10^-9 * 100

Q = 7.59*10^-7Coulombs

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Q = 0.759µC

b) Energy stored in a capacitor is expressed as E = 1/2CV²

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E = 39.5*10^-6Joules

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3 years ago
A 50-cm-long spring is suspended from the ceiling. A 270 g mass is connected to the end and held at rest with the spring unstret
I am Lyosha [343]

Answer:

The spring constant is 45.94 N/m.

Explanation:

Given that,

Length = 50 cm

Mass = 270 g

Stretching the spring = 24 cm

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The change in potential energy equal to the change in kinetic energy.

mgh=\dfrac{1}{2}kx^2

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270\times10^{-3}\times9.8\times50\times10^{-2}=\dfrac{1}{2}\times k\times(24\times10^{-2})^2

k=\dfrac{2\times270\times10^{-3}\times9.8\times50\times10^{-2}}{(24\times10^{-2})^2}

k=45.94\ N/m

Hence, The spring constant is 45.94 N/m.

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