What is the Mechanical advantage of the inclined plane? A) 0.25 B) 4 C) 90 D) 150

Find the equivalent resistance Req between terminals a and b if terminals c and d are open and again if terminals c and d are sh

adoni [48]

Answer:

351 ohm

720 ohm

Explanation:

When c and d are open:

Terminals c and d are open.. If you redraw the circuit as below, you can see that the two resistors in the first column are in parallel as, they are connected together at both pairs of terminals (due to the short).

Hence, we have a pair of parallel resistors:

Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms

Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms

Now these two sets are in series with another Hence,

Req = Req1 + Req2 = 216 + 135 = 351 ohms

Answer: 351 ohms

When c and d are shorted:

The current will flow through the least resistant path naturally from resistors R3 and R1 or R4.

Both of these resistor lie in a single path placing the resistors in series to one another, hence

Req = R3 + R1 = 180 + 540 = 720 ohms

Answer:720 ohms

7 0

3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

2 years ago

An empty oﬃce chair is at rest on a ﬂoor. Consider the following forces:. 1. A downward force due to gravity;. 2. An upward forc

Bumek [7]

4. 1 and 2 only.

1. the downward force is the force of gravity.

<span>2. The upward force exerted is the Normal reaction from the floor.</span>

8 0

3 years ago

Ashkon throws a basketball across the court to his teammate. The ball has 57 J of potential energy and 61 J of kinetic energy. W

Gwar [14]

The total mechanical energy of the ball is the sum of its potential energy U and its kinetic energy K, therefore:
$E=U+K=57 J+61 J=118 J$
so, the total mechanical energy of the basketball is 118 J.

8 0

3 years ago

Read 2 more answers

If two arm wrestlers exert a force on each other’s hands, and the hands don’t move, the forces must be *

jeyben [28]

Balanced. They’re equally as strong so as their arm wrestling, neither of the men’s hands go down. Because they’re equally/balanced as strong.

3 0

3 years ago