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3 years ago

Is it possible for a car to be accelerating to the west while it is driving to the east?

Physics

1 answer:

Ber [7]3 years ago

8 0

Answer:

Yes

Explanation:

If the acceleration has an opposite direction to the velocity of the car, this means that it is opposed to is motion. Therefore, it is called deceleration, since the car's velocity will decrease until it stops and then will start it moving towards the west.

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A pendulum has 711 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o

Anika [276]

According to law of conservation of energy,
Energy can neither be constructed nor be destroyed but can be transformed from one form to another.

At the highest point of the pendulum(point b), pendulum is associated with potential energy only and no kinetic energy.
Therefore total energy at point b = potential energy = 711 J.... i

At the bottom most point(point a), pendulum is associated only with kinetic energy and no potential energy.
Therefore total energy at point a = kinetic energy ---- ii
From i and ii,
Kinetic energy = potential energy = 711 J.(Conserving energy)

Hence kinetic energy at the bottom most point is 711 J.
Hope this helps!!

7 0

3 years ago

Which of the following is the best description of a gaseous substance?

sveta [45]

Answer:

It has no shape of its own but has a definite volume.

Explanation:

Gases have no shape but a definite volume

6 0

3 years ago

Read 2 more answers

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

3 years ago

An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc

Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

$P=VI$

Express this in terms of 'I'. This gives,

$I=\frac{P}{V}$

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, $P_1=16\ W, V=110\ V$

So, current drawn is given as:

$I_1=\frac{16\ W}{110\ V}=0.145\ A$

For incandescent bulb, $P_2=100\ W, V=110\ V$

So, current drawn is given as:

$I_2=\frac{100\ W}{110\ V}=0.909\ A$

For CFL bulb, $P_3=24\ W, V=110\ V$

So, current drawn is given as:

$I_3=\frac{24\ W}{110\ V}=0.218\ A$

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

5 0

3 years ago

An iron wire has length 8.0m and a diameter 0.50mm. The sir has a resistance R.

Rudik [331]

The resistance of the second wire is 16 R.
where R is the resistance of the first wire.

R = ρ $\frac{l}{A}$
where l = length of the wire
A = area of the wire
A = $\pi r^{2}$ where, r = $\frac{diameter of wire}{2}$

Thus, on finding the ratio of resistance of the two wires, we get,

$\frac{R1}{R2} = \frac{l1A2}{l2A1}$

here, R1 = R
l1 = 8m
l2 = 2m
A1=π $0.25^{2}$
A1=π $0.50^{2}$

we get. R2 = 16R

7 0

3 years ago