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3 years ago

7

Mass movement Choose one: A. depends on the difference between the downslope force and the resistance force. B. cannot happen un

der wet conditions. C. cannot happen underwater because the buoyancy force in water is too great. D. always happens when the slope of a hill is not as steep as its angle of repose.

1 answer:

STatiana [176]3 years ago

8 0

Answer: A

Depends on the difference between the downslope force and the resistance force.

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The amount of potential energy possessed by an elevated object is equal to

Sedbober [7]

Answer:

Potential energy of the object will be equal to mgh

Explanation:

Let the mass of the object is m

Acceleration due to gravity is $gm/sec^2$

Let the object is released from height h

We have to find the potential energy

Potential energy is of an object released from height h is equal to

$U=mgh$ , here m is mass, g is acceleration due to gravity and h is height from which object is released.

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4 years ago

Which scenarios are examples of physical change?

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Answer:

For example: Freezing, boiling, are physical

Explanation:

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A spherical balloon is made from a material whose mass is 3.00 kg. The thickness of the material is negligible compared to the 1

vesna_86 [32]

To solve the problem it is necessary to apply the definition of Newton's second Law and the definition of density.

Density means the relationship between volume and mass:

$\rho = \frac{m}{V}$

While Newton's second law expresses that force is given by

F = ma

Where,

m = mass

a= acceleration (gravity at this case)

In the case of the given data we have to,

$m_b = 3Kg$

$r = 1.5m\\V = \frac{4}{3}\pi r^3 \\V = \frac{4}{3} \pi 1.5^3\\V = 14.13m^3$

In equilibrium, the entire system is equal to zero, therefore

$\sum F = 0$

$F_g +F_h-F_b = 0$

Where,

$F_g =$ Weight of balloon

$F_h =$ Weight of helium gas

$F_b =$ Bouyant force

Then we have,

$mg+V\rho g -V\rho_a g = 0$

$\rho = \rho_0-\frac{m}{V}$

Replacing the values we have that

$\rho = 1.19kg/m^3 -\frac{3Kg}{14.13m^3}$

$\rho = 0.978kg/m^3$

Now by ideal gas law we have that

$PV=nRT$

$P\frac{\rho}{m} = nRT$

$P = \rho \frac{n}{m}RT$

But the relation \frac{n}{m} is equal to the inverse of molar mass, that is

$P = \frac{\rho}{M_0} RT$

$P = \frac{0.978kg/m^3}{0.04kg/mol}*8.314J/K.Mol * 305K$

$P = 619995.7Pa$

Therefore the pressure of the helium gas assuming it is ideal is 0.61Mpa

5 0

3 years ago

A high-speed K0 meson is traveling at β = 0.90 when it decays into a π + and a π − meson. What are the greatest and least speeds

san4es73 [151]

Answer:

greatest speed=0.99c

least speed=0.283c

Explanation:

To solve this problem, we have to go to frame of center of mass.

Total available energy fo π + and π - mesons will be difference in their rest energy:

$E_{0,K_{0} }-2E_{0,\pi } =497Mev-2*139.5Mev\\$

=218 Mev

now we have to assume that both meson have same kinetic energy so each will have K=109 Mev from following equation for kinetic energy we have,

K=(γ-1) $E_{0,\pi }$

$K=E_{0,\pi}(\frac{1}{\sqrt{1-\beta ^{2} } } -1)\\\frac{1}\sqrt{1-\beta ^{2} }}=\frac{K}{E_{0,\pi}}+1\\ {1-\beta ^{2}=\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta ^{2}=1-\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta = +-\sqrt{\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\\\\beta =+-\sqrt{1-\frac{1}{(\frac{109Mev}{139.5Mev+1)^2}}$

$u'=+-0.283c$

note +-=±

To find speed least and greatest speed of meson we would use relativistic velocity addition equations:

$u=\frac{u'+v}{1+\frac{v}{c^{2} } } u'\\u_{max} =\frac{u'_{+} +v_{} }{1+\frac{v}{c^{2} } } u'_{+} \\u_{max} =\frac{0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } 0.828\\ u_{max} =0.99c\\u_{min} =\frac{u'_{-} +v_{} }{1+\frac{v}{c^{2} } } u'_{-}\\u_{min} =\frac{-0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } -0.828c\\u_{min} =0.283c$

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3 years ago

A box of mass 13kg sits on a table with a coefficient of static friction of 0.85. What is the maximum force of static friction?

Alexus [3.1K]

Answer:

108.3 N

Explanation:

The maximum force of static friction acting on the box is given by:

$F=\mu N$

where

$\mu = 0.85$ is the coefficient of static friction

N is the normal reaction of the table on the box

Since the box is in equilibrium along the vertical direction, the normal reaction N is equal to the weight of the box, so:

$N=mg=(13 kg)(9.8 m/s^2)=127.4 N$

And so, the maximum force of static friction is

$F=(0.85)(127.4 N)=108.3 N$

7 0

3 years ago