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3 years ago

What is the amplitude of oscillation a of the scale after the slices of ham land on the plate?

Physics

1 answer:

fgiga [73]3 years ago

4 0

Answer:

0.093

Explanation:

A = √(mgh/K)

M= 0.7 h= 0.25

K= 200 g= 9.8

= √(0.7*9.8*0.25)/200

= √(1.715)/200

=√0.008575

A = 0.093m

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Gallium arsenide (GaAs) and gallium phosphide (GaP) are compound semiconductors that have room-temperature band gap energies of

Firdavs [7]

Hence, 0.60+2.25eV is the solution to this problem

3 0

2 years ago

The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig

Virty [35]

Answer:

1. The tension in the rope is everywhere the same.

2. The magnitudes of the forces exerted on the two objects by the rope are the same.

3. The forces exerted on the two objects by the rope must be in opposite directions.

Explanation:

"Massless ropes" do not have a "net force" which means that it is able to transmit the force from one end of the rope to the other end, perfectly. It is known for its property of having a total force of zero. In order to attain this property, the magnitude of the forces exerted on the two stationary objects by the rope are the same and in opposite direction. So this explains number 2 & 3 answers.

Since the objects that are held by the rope are stationary, then this means that the tension in the rope is also stationary. This means that the tension in the rope everywhere is the same (provided that the rope is still or in a straight line, as stated in the situation above, and is being held by two points). So, this explains number 1.

6 0

3 years ago

A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m verticall

lorasvet [3.4K]

Explanation:

Given:

t = 20 seconds

x = 3000 m

y = 450 m

a) To find the vertical component of the initial velocity $v_{0y}$ , we can use the equation

$y = v_{0y}t - \frac{1}{2}gt^2$

Solving for $v_{0y}$ ,

$v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}$

$\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}$

$\:\:\:\:\:\:\:=120.5\:\text{m/s}$

b) We can solve for the horizontal component of the velocity $v_{0x}$ as

$x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}$

$v_{0x} = 150\:\text{m/s}$

4 0

3 years ago

Car travels 40km/hr for half an hour. How far does it travel

Lerok [7]

Answer:

The car traveled, d = 20 Km

Explanation:

Given that,

The speed of the car, v = 40 Km/h

The time taken by the car is, t = 0.5 h

The distance covered by the car for half an hour, d = ?

The relation between the velocity and displacement is given by the formula,

v = d / t

Therefore,

d = v x t

Substituting the given values in the above equation,

d = 40 Km/h x 0.5 h

= 20 Km

Hence, the distance covered by the car is, d = 10 Km

7 0

3 years ago

A motorcycle traveling at 36 m/s slams on the brakes to avoid an accident. The motorcycle skids 23m before stoping. What is the

Svetach [21]

Since the motor cycle was moving at 36 m/s before the brakes were applied, the initial velocity
$u = 36 {ms}^{ - 1}$
The motorcycle traveled 23m before stopping, that means
$s = 23m$
Since the motorcycle stopped moving the final velocity is zero.

$v = 0{ms}^{ - 1}$
We can use this 'suvat' equation of linear motion.

${v}^{2} = {u}^{2} + 2as$
Plugging in the above values gives,

${0}^{2} = {36}^{2} + 2a(23)$

$0 = 1296+ 46a$
$46a = - 1296$
$a = - 28.2 {ms}^{ - 2}$

We can use the equation
$v = u + at$
to find the time the motorcycle took to stop.

$0 = 36 + - 28.2t$
$- 36 = - 28.2t$
$t = 1.3s$

8 0

3 years ago