All categories

1 year ago

A simple arrangement by means of which e.m.f,s. are compared is known

Physics

1 answer:

strojnjashka [21]1 year ago

5 0

Answer:

A simple arrangement by means of which e.m.f,s. are compared is known as?

(a)Voltmeter

(b)Potentiometer

(c)Ammeter

(d)None of the above

Explanation:

You might be interested in

At a local swimming pool, the diving board is elevated h = 5.5 m above the pool's surface and overhangs the pool edge by L = 2 m

Margaret [11]

Answer:

Part a)

$t = \sqrt{\frac{2h}{g}}$

Part b)

$t = 1.06 s$

Part c)

$L = 4.86 m$

Explanation:

Part a)

The height of the diving board is given as

$h = 5.5 m$

now the speed of the diver is given as

$v_0 = 2.7 m/s$

when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board

So we will have

$y = v_y t + \frac{1}{2}at^2$

$h = 0 + \frac{1}{2}gt^2$

$t = \sqrt{\frac{2h}{g}}$

Part b)

$t = \sqrt{\frac{2h}{g}}$

plug in the values in the above equation

$t = \sqrt{\frac{2(5.5 m)}{9.81}$

$t = 1.06 s$

Part c)

Horizontal distance moved by the diver is given as

$d = v_0 t$

$d = 2.7 \times 1.06$

$d = 2.86 m$

so the distance from the edge of the pool is given as

$L = 2.86 + 2$

$L = 4.86 m$

4 0

1 year ago

The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a

navik [9.2K]

Answer:

230

Explanation:

$\omega$ = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

$K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J$

Energy used in one acceleration

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J$

Number of accelerations would be given by

$n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4$

So the number of complete accelerations is 230

8 0

2 years ago

A professor's office door is 0.99 m wide, 2.2 m high, 4.2 cm thick; has a mass of 27 kg, and pivots on frictionless hinges. A "d

ANEK [815]

Answer:

$I=8.8209\ kg.m^2$

$\alpha=0.6348\ rad.s^{-2}$

Explanation:

Given:

width of door, $w=0.99\ m$

height of the door, $h=2.2\ m$

thickness of the door, $t=4.2\ cm$

mass of the door, $m=27\ kg$

torque on the door, $\tau=5.6\ N.m$

<em>∵Since the thickness of the door is very less as compared to its other dimensions, therefore we treat it as a rectangular sheet.</em>

For a rectangular sheet we have the mass moment of inertia inertia as:

$I=\frac{1}{3} m.w^2$

$I=\frac{1}{3}\times 27\times 0.99^2$

$I=8.8209\ kg.m^2$

We have a relation between mass moment of inertia, torque and angular acceleration as:

$\alpha=\frac{\tau}{I}$

$\alpha=\frac{5.6}{8.8209}$

$\alpha=0.6348\ rad.s^{-2}$

6 0

1 year ago

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

SCORPION-xisa [38]

Answer:

The value is $r = 5.077 \ m$

Explanation:

From the question we are told that

The Coulomb constant is $k = 9.0 *10^{9} \ N\cdot m^2 /C^2$

The charge on the electron/proton is $e = 1.6*10^{-19} \ C$

The mass of proton $m_{proton} = 1.67*10^{-27} \ kg$

The mass of electron is $m_{electron } = 9.11 *10^{-31} \ kg$

Generally for the electron to be held up by the force gravity

Then

Electric force on the electron = The gravitational Force

i.e

$m_{electron} * g = \frac{ k * e^2 }{r^2 }$

$\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81$

$r = \sqrt{25.78}$

$r = 5.077 \ m$

7 0

1 year ago

A ball is thrown with a velocity of 35 meters per second at an angle of 30° above the horizontal. which quantity has a magnitude

enot [183]

The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is the vertical velocity.

In fact, the motion of the ball consists of two separate motions:
- the horizontal motion, on the x-axis, which is a uniform motion with constant velocity $v_x=v_0 cos 30^{\circ}$ , where $v_0=35 m/s$
- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ directed downwards, and with initial velocity $v_y=v_= sin 30^{\circ}$ . Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height.

5 0

1 year ago