All categories

2 years ago

What happens to the particles of a liquid when energy is removed from them?

Physics

1 answer:

KonstantinChe [14]2 years ago

3 0

Answer:

D: The distance between the particles decreases

Explanation:

Taking away energy slows down molecules, like how you slow down when you are cold (I think)

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

2 years ago

A disk is spinning about its center with a constant angular speed at first. Let the turntable spin faster and faster, with const

hoa [83]

Answer:

4 (please see the attached file)

Explanation:

While the angular speed (counterclockwise) remained constant, the angular acceleration was just zero.

So, the only force acting on the bug (parallel to the surface) was the centripetal force, producing a centripetal acceleration directed towards the center of the disk.

When the turntable started to spin faster and faster, this caused a change in the angular speed, represented by the appearance of an angular acceleration α.

This acceleration is related with the tangential acceleration, by this expression:

at = α*r

This acceleration, tangent to the disk (aiming in the same direction of the movement, which is counterclockwise, as showed in the pictures) adds vectorially with the centripetal force, giving a resultant like the one showed in the sketch Nº 4.

7 0

2 years ago

Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and

yKpoI14uk [10]

Answer:

The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

$v=\dfrac{dx}{dt}$ ....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

$a=\dfrac{dv}{dt}$ ....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

$F=\dfrac{d(p)}{dt}$

Put the value of p

$F=\dfrac{d(mv)}{dt}$

$F=m\dfrac{dv}{dt}$

Put the value from equation (II)

$F=ma$

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

$F=\dfrac{Gm_{1}m_{2}}{r^2}$

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

3 0

2 years ago

Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward

KATRIN_1 [288]

Answer:

Tension in the cable is T = 16653.32 N

Explanation:

Give data:

Cross section Area A = 1.3 m^2

Drag coefficient CD = 1.2

Velocity V = 4.3 m/s

Angle made by cable with horizontal =30 degree

Density $\rho \ of\ water= 1000 kg/m3$

Drag force FD is given as

$F_{D} = \fracP{1}{2} \rho v^{2} C_{D} A$

$= 0.5\times 1000\times 4.32\times 1.2\times 1.3$

Drag force = 14422.2 N acting opposite to the motion

As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force

TCos30 = F_D

$T = \frac{F_D}{cos30}$

$T =\frac{ 14422.2}{cos 30}$

T = 16653.32 N

7 0

2 years ago

An x-ray tube is an evacuated glass tube that produces electrons at one end and then accelerates them to very high speeds by the

laila [671]

Answer:

a) V = 1.866 10² V , b) V = 3.424 10⁵ V , c) v = 8.1 10⁶ m / s

Explanation:

a) the potential difference is requested to accelerate the electrons up to 2.7% of the speed of light

v = 0.027 c

v = 0.027 3 10⁸

v = 8.1 10⁶ m / s

for this part we can use the conservation of mechanical energy

starting point. When electrons are at rest

Em₀ = U = q V

final point. Electrons with maximum speed

Em_f = K = ½ m v2

Em₀ = Em_{f}

e V = ½ m v²

V = ½ m v² / e

let's calculate

V = ½ 9.1 10⁻³¹ (8.1 10⁶)² / 1.6 10⁻¹⁹

V = 1.866 10² V

V = 1866 V

b) if this acceleration protons is the mass of the proton is m_{p} = 1.67 10-27

V = ½ 1.67 10⁻²⁷ (8.1 10⁶)² / 1.6 10⁻¹⁹

V = 3.424 10⁵ V

V = 342402 V

c) this potential difference should give the protons the same speed as the electrons

v = 8.1 10⁶ m / s

5 0

2 years ago