Ask question

Ask question

All categories

3 years ago

11

PLZ HELP ASAP! 2.03 QUIZ: TEMPERATURE AND HEAT

1 answer:

gladu [14]3 years ago

7 0

A. Its A for sure but I don't like the way it's written here, they didn't get the physics right

The temperature increases BECAUSE the kinetic energy of the molecules increases

You might be interested in

A wheel of radius 0.23 m, which is moving initially at 25.0 m/s, rolls to a stop in 246.0 m. The wheel's rotational inertia is 0

Digiron [165]

Answer: -1.27 m/s^2

Explanation:

a = - V^2 / 2x

a = -(25^2) / 2 x (246) = 1.27 m/ s^2

Therefore the linear acceleration of the wheel is - 1.27 m/s^2

8 0

3 years ago

Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.)

Taya2010 [7]

Answer:

(a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

$P_{1}=\rho g h_{1}$

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

$P_{1}=\rho gh_{1}$ .....(I)

Pressure for second pipe,

$P_{2}=\rho gh_{2}$ .....(II)

From equation (I) and (II)

$P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

Put the value of P₁ and P₂

$\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)$

$2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2$ ....(III)

We know that,

The continuity equation

$v_{1}A_{1}=v_{2}A_{2}$

$v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})$

Put the value of v₂ in equation (III)

$2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2$

$2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2$

Here, $\dfrac{A_{1}}{A_{2}}=\gamma$

So, $2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)$

$v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Hence, (a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

8 0

3 years ago

A small block of mass 1.57 kg rests on the left edge of a block of length 3.45 m and mass 7.04 kg. the coefficient of kinetic fr

MariettaO [177]

...............................................................

3 0

3 years ago

Fish give off the compound ammonia, which has a pH above 7. To which class of compounds does ammonia belong?

Amanda [17]

It belongs to hydrides, it’s a nitrogen hydride
It is a base because of pH and because it’s a soluble vase it is an alkali

4 0

3 years ago

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total

Harlamova29_29 [7]

Answer:

a)0.48 m/s

b) 0.583 m/s

Explanation:

As the wagon rolls,

momentum'p'= m x v => 95.8 x 0.530 = 50.774 Kgm/s

(a)Rock is thrown forward,

momentum of rock = 0.325 x 15.1 = 4.9075 Kgm/s

Conservation of momentum says momentum of wagon is given by

50.774 - 4.9075 = 45.8665

Therefore, Speed of wagon = 45.8665 / (95.8-0.325) = 0.48 m/s

(b) Rock is thrown backward,

momentum of wagon = 50.774 + 4.9075 = 55.68 Kgm/s

Therefore, speed of wagon = 55.68 / (95.8-0.325) = 0.583 m/s

4 0

4 years ago