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3 years ago

If a car is traveling eastward, can its acceleration be westward?

2 answers:

5 0

Yes it can. It may be a bit counterintuitive but if the car is slowing down, even if it's traveling east, its acceleration is toward the west because it's decelerating and slowing down.

gizmo_the_mogwai [7]3 years ago

4 0

The motions moving in opposite directions cancel each other when they have the same magnitude or speed in this case. If the car travels towards east, then it can be possible that the acceleration goes west as this case points to deceleration. The acceleration then has a negative sign.

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All of the following are physical stimuli for touch except __________ energy.

hodyreva [135]

C. combustion I think. hope this helps

6 0

3 years ago

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The centripetal force acting on a satellite in orbit

krok68 [10]

The centripetal force (of gravity) on a satellite in orbit is an
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the satellite away from Earth), changes the direction of the
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3 years ago

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Consider a series RLC circuit where R = 855 Ω and C = 6.25 μF. However, the inductance L of the inductor is unknown. To find its

sashaice [31]

Answer:

L= 0.059 mH

Explanation:

Given that

R = 855 Ω and C = 6.25 μF

V= 84 V

Frequency

ω = 51900 1/s

We know that

$\omega=\sqrt{\dfrac{1}{LC}}$

L=Inductance

C=Capacitance

ω =angular Frequency

ω² L C =1

(51900)² x L x 6.25 x 10⁻⁶ = 1

L= 5.99 x 10⁻⁵ H

L= 0.059 mH

6 0

3 years ago

A battery is connected to a light bulb that has 3 of resistance . If there is 0.5 A of current flowing what is the voltage on th

Alenkasestr [34]

Answer:

1.5 V

Explanation:

E = IR = 0.5(3) = 1.5 V

8 0

3 years ago

The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc

qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

$R = \frac{6}{2.5} - 2.4 ohm$

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

$\frac{3R}{R+3} \times 2.5 = 6$

R= 12 ohm

2) circuit Q

$I\times (R+0.1) =V$

$R+0.1 =\frac{6}{2.5}$

R = 2.3 ohm

5 0

3 years ago