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2 years ago

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Which statement is true of forces?

1 answer:

jasenka [17]2 years ago

8 0

FORCES HAVE STRENGTH AND DIRECTION

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An 8.5-l tire is filled with 0.55 mol of gas at a temperature of 305 k. what is the pressure of the gas in the tire?

const2013 [10]

Answer:

1.62 atm

Explanation:

We can solve the problem by using the ideal gas equation:

$pV=nRT$

where:

p = ? is the pressure of the gas in the tire

V = 8.5 L is the volume of the tire

n = 0.55 mol is the number of moles of the gas

R = 0.0821 atm L / K mol is the gas constant

T = 305 K is the temperature of the gas

By re-arranging the equation and substituting the numbers in, we find:

$p=\frac{nRT}{V}=\frac{(0.55 mol)(0.0821 Latm/Kmol)(305 K)}{8.5 L}=1.62 atm$

8 0

3 years ago

What is Newton's law of universal gravitation?

pshichka [43]

Newton's law of universal gravitation<span> states that a particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. hope this helps =)</span>

8 0

2 years ago

Read 2 more answers

An object is dropped from an altitude of one Earth radiusabove Earth's surface. If M is the mass of Earth and R is itsradius the

Montano1993 [528]

Answer:

The speed of the object just before it hits Earth is $\sqrt{\dfrac{GM}{R}}$

(A) is correct option.

Explanation:

Given that,

M = mass of earth

R = radius of earth

The potential energy at height above the surface of the earth

$P.E=-\dfrac{GmM}{R+h}$

The kinetic energy at height above the surface of the earth

$K.E = 0$

The total energy at height above the surface of the earth

$E = K.E+P.E$

$E = -\dfrac{GmM}{R+h}$ ....(I)

The total energy at the surface of the earth

$E'=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$ ....(II)

We need to calculate the speed of the object just before it hits Earth

From equation (I) and (II)

$-\dfrac{GmM}{R+h}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$

Here, h = R

$-\dfrac{GmM}{2R}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$

$v= \sqrt{\dfrac{GM}{R}}$

Hence, The speed of the object just before it hits Earth is $\sqrt{\dfrac{GM}{R}}$ .

4 0

2 years ago

The distance between the crest and trough of an ocean wave is 1 meter. What is the amplitude of the wave?

Darya [45]

Answer:

distance between crust and trough = 1m

amplitude = 1/2 = 0.5

8 0

1 year ago

A skier of mass 72 kg is pulled up a slope by a motor-driven cable. a. how much work is required to pull him 75 m up a 30 degree

likoan [24]

Answer: $26.460\times 10^3\ J$

Explanation:

Given the mass of skier m=72 kg

distance traveled d=75 m

constant speed v=3.4 m/s

If speed is constant then there must no force acting in the direction of motion

i.e. tension force must be equal to the component of weight

$T=mg\sin 30^{\circ}$

Work done is given by

$\Rightarrow W=F\cdot d=Td\cos 0^{\circ}\\\Rightarrow W=mg\sin 30^{\circ} d=72\times 9.8\times 0.5\times 75\\\Rightarrow W=26.460\times 10^3\ J$

8 0

2 years ago