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1 year ago

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A simple pendulum is suspended from the ceiling of a stationary elevator, and the period is determined.(i) When the elevator acc

elerates upward, is the period (a) greater(b) smaller(c) unchanged?

1 answer:

ELEN [110]1 year ago

6 0

The correct option is (a) greater.

<h3>What is simple pendulum?</h3>

A weight hanging from a pivot such that it may freely swing is called a pendulum. Gravity's restoring force will cause a pendulum to accelerate back toward its equilibrium position if it is sideways moved from its resting, equilibrium position.

The effective acceleration of the pendulum is equal to the sum of the acceleration caused by gravity and the elevator's upward acceleration while the elevator is moving upward.

T=2π√L/g(eff)

T=2π√L/(g+a)

Therefore, effective acceleration increases, and which results in a decrease in the period of motion.

to learn more about simple pendulum go to -

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A man stands by a train track to watch the train car go by. Inside, a man walks through the train car eating a hot dog (the hot

antiseptic1488 [7]

Answer:

D) 11 m/s

Explanation:

The problem asks us to calculate the velocity of the hot dog with respect to the observer stationary outside the train. This velocity is given by:

$v=v_t + v_m + v_h$

where

$v_t=+10 m/s$ is the velocity of the train (towards right)

$v_m=+2 m/s$ is the velocity of the man (towards right)

$v_h=-1 m/s$ is the velocity of the hot-dog (towards left, so we put a negative sign)

Substituting the numbers into the equation, we find

$v=+10 m/s+2 m/s-1 m/s=+11 m/s$

and the positive sign means the velocity is toward right.

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2 years ago

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When an unbalanced force acts on an object the change in the objects ____or ____ depends on the size and direction of the force

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When an unbalanced force acts on an object the change in the object state of rest or motion depends on the size and direction of the force.

If a body is at state of rest or motion, when an unbalanced external force acts on it, its starts moving in the direction of force and magnitude of its velocity or acceleration depends on the magnitude of force applied.

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3 years ago

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The beginning of the Phanerozoic is marked by what occurrence

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The beginning of the Phanerozoic is marked by the development of hard body parts, such as shells and bones.

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3 years ago

In a certain time period a coil of wire is rotated from one orientation to another with respect to a uniform 0.38-T magnetic fie

Evgesh-ka [11]

Answer:

8.4 V

Explanation:

induced emf, e1 = 5.8 V

Magnetic field, B1 = 0.38 T

magnetic field, B2 = 0.55 T

induced emf, e2 = ?

As we know that the induced emf is directly proportional to the magnetic field strength.

When the other parameters remains constant then

$\frac{e_{1}}{e_{2}}=\frac{B_{1}}{B_{2}}$

$\frac{5.8}{e_{2}}=\frac{0.38}{0.55}$

e2 = 8.4 V

Thus, the induced emf is 8.4 V.

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2 years ago

A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the

DiKsa [7]

Answer:

a). M = 20.392 kg

b). am = 0.56 $m/s^2$ (block), aM = 0.28 $m/s^2$ (bucket)

Explanation:

a). We got N = mg cos θ,

f = $\mu_s N$

= $\mu_s mg \cos \theta$

If the block is ready to slide,

T = mg sin θ + f

T = mg sin θ + $\mu_s mg \cos \theta$ .....(i)

2T = Mg ..........(ii)

Putting (ii) in (i), we get

$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$

$M=2(m \sin \theta + \mu_s mg \cos \theta)$

$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$

M = 20.392 kg

b). $(h-x_m)+(h-x_M)+(h'+x_M)=l$ .............(iii)

Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

$-\ddot{x}-2\ddot x_M=0$

$\ddot x_M=\frac{\ddot x_m}{2}$

$a_M=\frac{a_m}{2}$ .....................(iv)

We got, N = mg cos θ

$f_K=\mu_K mg \cos \theta$

∴ $T-(mg \sin \theta + f_K) = ma_m$

$T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$ ................(v)

Mg - 2T = M $a_M$

$Mg-Ma_M=2T$

$Mg-\frac{Ma_M}{2} = 2T$ (from equation (iv))

$\frac{Mg}{2}-\frac{Ma_M}{4}=T$ .....................(vi)

Putting (vi) in equation (v),

$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$

$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$

$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$

$a_m= 0.56 \ m/s^2$

Using equation (iv), we get,

$a_M= 0.28 \ m/s^2$

6 0

2 years ago