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3 years ago

Read about Pendulums. Then complete the sentences based on the diagram of the pendulum. Be sure to keep this

Physics

2 answers:

marin [14]3 years ago

5 0

Answer:

The pendulum has the most potential energy .

The pendulum has the least kinetic energy .

The pendulum has the most kinetic energy .

The pendulum has the least potential energy .

1) at the top of the swing

2) at the top of the swing

3) at the bottom of the swing

4) at the bottom of the swing

Explanation:

just did it, it's right. (EDGE 2020)

MrRissso [65]3 years ago

3 0

Answer:

Potential Energy is given as:

P.E = mgh

which means P.E is directly proportional to height 'h'.

Kinetic Energy is given as:

K.E = (1/2)mv²

which means K.E is directly proportional to velocity 'v²'.

Total Energy of Pendulum = K.E + P.E

1. The pendulum has the most potential energy at extreme position because the height is maximum at the extreme position.

2. The pendulum has the least kinetic energy at extreme position because the velocity is zero at extreme position.

3. The pendulum has most kinetic energy at the mean position because the velocity is maximum at this point.

4. The pendulum has the least potential energy at the mean position because the height is minimum.

A rule to remember: The point where K.E is maximum, P.E is zero at this point and vice versa.

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A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction

yawa3891 [41]

Answer:

$F'=\dfrac{F}{4}$

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

$F=\dfrac{mv^2}{r}$

v = 2v

$F=\dfrac{m(2v)^2}{r}$

$F=4\dfrac{m(v)^2}{r}$ ..........(1)

For the slower car on the road,

$F'=\dfrac{mv^2}{r}$ ............(2)

Equation (1) becomes,

$F=4F'$

$F'=\dfrac{F}{4}$

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

8 0

3 years ago

HELPPP PLEASE !!!!!!!

vlabodo [156]

Explanation:

a) d = ½.a.t²

200 = ½(4)t²

200 = 2t²

t² = 200/2

t² = 100

t =√100 = 10 s

b) Vt = a. t

= 4(10)

= 40 m/s

c) V av. = d/t = 200/10 = 20m/s

6 0

2 years ago

A student walks 4 blocks east, 7 blocks west, 1 block east and then 2 blocks west in an hour what is their velocity

Nutka1998 [239]

Answer:

4 blocks west is final displacement. So 4 blocks per hour

3 0

2 years ago

A 40.0-mH inductor is connected to a North American electrical outlet (ΔVrms = 120 V, f = 60.0 Hz). Assuming the energy stored i

trasher [3.6K]

Explanation:

It is given that,

Inductance, $L=40\ mH=40\times 10^{-3}\ H$

RMS value of voltage, $v_{rms}=120\ V$

Frequency, f = 60 Hz

We need to find the energy stored at t = (1 /185) s. It is assumed that energy stored in the inductor is zero at t = 0. So,

The current flowing through the inductor is given by :

$I_t=\dfrac{V_o}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})$

$I_t=\dfrac{\sqrt{2} V_{rms}}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt{2}}{2\pi f L}\ sin(2\pi f t-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt{2}}{2\pi\times 60\times 40\times 10^{-3}}\ sin(2\pi \times 60\times \dfrac{1}{185})-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt2}{15.07}\ sin(2\pi \times 60\times \dfrac{1}{185}-\dfrac{\pi}{2})$

I = 0.091 A

Energy stored in the inductor is, $U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times 40\times 10^{-3}\times (0.091)^2$

U = 0.000165 Joules

Hence, this is the required solution.

6 0

3 years ago

Calculate the kinetic energy in joules of an automobile weighing 2135 lb and traveling at 55 mph. (1 mile = 1.6093 km, 1 lb = 45

victus00 [196]

<span>Let's convert the speed to m/s: speed = (55 mph) (1609.3 m / mile) (1 hour / 3600 seconds) speed = 24.59 m/s Let's convert the mass to kilograms: mass = (2135 lb) (0.45359 kg / lb) mass = 968.4 kg We can find the kinetic energy KE: KE = (1/2) m v^2 KE = (1/2) (968.4 kg) (24.59 m/s)^2 KE = 292780 joules The kinetic energy of the automobile is 292780 joules.</span>

4 0

3 years ago