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2 years ago

Help (๑¯◡¯๑)me please

Physics

2 answers:

Allushta [10]2 years ago

4 0

Answer:

<h3>Young modulus of elasticity for a gas is</h3><h2>Zero</h2>

Explanation:

As the gas doesn't undergo any changes 

so the young modules of gas is not defined...

leonid [27]2 years ago

3 0

We know

$\\ \rm\hookrightarrow Y=\dfrac{Stress}{Strain}$

$\\ \rm\hookrightarrow Y=\dfrac{\sigma}{\Delta \ell/\ell}$

Stress is zero

$\\ \rm\hookrightarrow Y=\dfrac{0}{\Delta \ell/\ell}$

$\\ \rm\hookrightarrow Y=0$

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A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The

VARVARA [1.3K]

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25

using conservation of energy

$\dfrac{1}{2}mu^2 = m g d sin \theta$

$u^2 = 2 g d sin30^0$

$u= \sqrt{2\times 9.8 \times 10 sin30^0}$

u = 9.89 m/s

Using second law of motion

ma =μ mg

a = μ g

a = 0.25 x 9.8

a = 2.45 m/s²

Using third equation of motion ,

v² - u² = 2 a s

0² - 9.89² = 2 x 2.45 x s

s = 20 m

the distance moved before stopping is 20 m

3 0

3 years ago

A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

$\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

6 0

3 years ago

Read 2 more answers

You're driving in a car at 50 km/h and bump into a car ahead traveling at 48 km/h in the same direction. the speed of impact is

salantis [7]

To solve this problem, we must remember about the law of conservation of momentum. The initial momentum mist be equal to the final momentum, that is:

m1 v1 + m2 v2 = (m1 + m2) v’

where v’ is the speed of impact

Since we are not given the masses of each car m1 and m2, so let us assume that they are equal, such that:

m1 = m2 = m

Which makes the equation:

m v1 + m v2 = (2 m) v’

Cancelling m and substituting the v values:

50 + 48 = 2 v’

2 v’ = 98

v ‘ = 49 km/h

The speed of impact is 49 km/h.

6 0

3 years ago

Which country had the largest population in 1997

nignag [31]

China i hope this helped

6 0

3 years ago

) A striker can give the ball an initial speed of 30m/s. Within what two elevation angles must he kick

sveticcg [70]

Answer:

about 19.6° and 73.2°

Explanation:

The equation for ballistic motion in Cartesian coordinates for some launch angle α can be written ...

y = -4.9(x/s·sec(α))² +x·tan(α)

where s is the launch speed in meters per second.

We want y=2.44 for x=50, so this resolves to a quadratic equation in tan(α):

-13.6111·tan(α)² +50·tan(α) -16.0511 = 0

This has solutions ...

tan(α) = 0.355408 or 3.31806

The corresponding angles are ...

α = 19.5656° or 73.2282°

The elevation angle must lie between 19.6° and 73.2° for the ball to score a goal.

_____

I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.

5 0

3 years ago

Help (๑¯◡¯๑)me please ​

Help (๑¯◡¯๑)me please