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3 years ago

CAN SOMEONE PLEASE HELP ME!!? I NEED HELP PLEASE!!

Physics

1 answer:

Dmitry [639]3 years ago

3 0

The first one is decreases because there are less molecules in the air.
The second on would be the last option, August 29, because a new moon occurs every 29 days.
The answer for the last one would be A, I presume, because when air masses arrive at a destination, it becomes dry.

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A roller coaster car is going over the top of a 18-mm-radius circular rise. At the top of the hill, the passengers "feel light,"

Andre45 [30]

Answer:

0.29713 m/s

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity

r = Radius = 18 mm

By balancing the forces in the system we have

$mg-N=\dfrac{mv^2}{r}\\\Rightarrow mg-\dfrac{mg}{2}=\dfrac{mv^2}{r}\\\Rightarrow v=\sqrt{r(g-\dfrac{g}{2})}\\\Rightarrow v=\sqrt{0.018\times (9.81-\dfrac{9.81}{2})}\\\Rightarrow v=0.29713\ m/s$

The velocity of the coaster is 0.29713 m/s

7 0

3 years ago

Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?

Alexandra [31]

11) $1.04\cdot 10^7 J$

12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

where

$KE_t$ is the kinetic energy at the top

$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

$PE_b=0$

Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s$

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

where

$KE_1$ is the kinetic energy at the top of the 1st hill

$PE_1$ is the potential energy at the top of the 1st hill

$KE_2$ is the kinetic energy at the top of the 2nd hill

$PE_2$ is the potential energy at the top of the 2nd hill

From the previous questions, we know that

$KE_1=0$

and

$PE_1=1.04\cdot 10^7 J$

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

$PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J$

So, the kinetic energy at the second hill is

$KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J$

And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

4 0

3 years ago

What are two ways in which all types of precipitation are alike

Digiron [165]

1. they all come from the clouds 2.they are all formed by precipatation. hope this helps enjoyyyyyy!!!!!!!!!!!!!

7 0

3 years ago

A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x ax

alexdok [17]

Answer:

Explanation:

We shall apply conservation of momentum law in vector form to solve the problem .

Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

= .444 i

momentum of 22 g

= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

applying conservation law of momentum

p + .444 i + .748 j = 0

p = - .444 i - .748 j

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

= 24 g = .024 kg

if v be its velocity

.024 v = .87

v = 36.25 m / s .

6 0

3 years ago

10. A loud motorcycle passes by the front of Bill's house. What medium

garik1379 [7]

Answer: Air

Explanation:

3 0

3 years ago

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