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MrRa [10]
3 years ago
12

Juliana walks 45 meters East, 45 meters south and 45 meters north. What is her resultant displacement?

Physics
1 answer:
katrin2010 [14]3 years ago
6 0

Her resultant displacement is (45 Meters East.)

She originally walks 45 meters east, then she goes south 45 meters, then north 45 meters.  The south and north placements  just make her go back to where her previous placements were before them. Making her 45 meters east.

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Choose true or false for each statement regarding a parallel plate capacitor.. The voltage of a disconnected charged capacitor i
OlgaM077 [116]

Answer:

Explanation:

The voltage of a disconnected charged capacitor increases when the plate area is decreased.

When plate area decreases , capacitance C decreases , but charge Q remains constant .

Q = C V where C is capacitance and V is voltage .

when C decreases , V increases for keeping Q constant .

So the statement is true.

The electric field is dependent on the charge density on the plates.

This statement is true .

The voltage of a connected charged capacitor remains the same when the plate area is decreased .

For a connected capacitor , V or voltage is constant which is equal to voltage of charging battery .

So the statement is true .

3 0
3 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
Plz help..During takeoff a plane accelerates at 4m/s^2 and takes 40s to reach takeoff speed.
matrenka [14]

Answer:

The velocity of the plane at take off is 160 m/s.

The distance travel by the plane in that time is 3200 meter.

Explanation:

Given:

Acceleration, a = 4 m/s²

Time, t = 40 s

u = 0 i .e initial velocity

To Find:

velocity , v = ?

distance , s =?

Solution:

we have first Kinematic equation

v = u + at

∴ v = 0 + 4×40

∴ v = 160 m/s

Now by Third Kinematic equation

s = ut + \frac{1}{2}at^{2}

∴ s = 0 + 0.5 × 4× 40²

∴ s = 3200 meter

3 0
2 years ago
Name 2 different "limiting Factors" that limit the size of a population in a given ecosystem.
Scrat [10]

Answer:    food, water, habitat, and mate.

Explanation:      The common limiting factors in an ecosystem are food, water, habitat, and mate. The availability of these factors will affect the carrying capacity of an environment. As population increases, food demand increases as well

8 0
2 years ago
Suppose we measure the energy stored in some inductor to be E when there is a current I running through it. If I double the curr
slavikrds [6]

Answer:

If I double the current in the inductor, the new total energy will become 4E (option f).

Explanation:

The coil or inductor is a passive component made of an insulated wire that stores energy in the form of a magnetic field due to its form of coiled turns of wire, through a phenomenon called self-induction. In other words, inductors store energy in the form of a magnetic field. The energy stored in the space where there is a magnetic field in the inductor is:

E=\frac{1}{2} *L*I^{2}

where E is Energy [J], L is Inductance [H] and I is Current [A].

If you double the current in the inductor, then the new value of the current is I'= 2*I. So replacing the new total energy is:

E'=\frac{1}{2} *L*I'^{2}=\frac{1}{2} *L*(2*I)^{2}=\frac{1}{2} *L*4*I^{2}=4*\frac{1}{2} *L*I^{2}

Then:

E'=4*E

<em><u>If I double the current in the inductor, the new total energy will become 4E (option f).</u></em>

3 0
2 years ago
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