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2 years ago

Check my work please

1 answer:

5 0

We can use the ideal gas equation which is expressed as PV = nRT. At a constant volume and number of moles of the gas the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:

T1/P1 = T2/P2

P2 = T2 x P1 / T1

P2 = 273 x 340 / 713

P2 = 130 kPa

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Which is a characteristic of diatoms?

zlopas [31]

Answer:

Explanation:

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3 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is 0.833.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

3 years ago

On a horizontal surface is located

Ierofanga [76]

By Newton's second law, the net vertical force acting on the object is 0, so that

n - w = 0

where n = magnitude of the normal force of the surface pushing up on the object, and w = weight of the object. Hence n = w = mg = 196 N, where m = 20 kg and g = 9.80 m/s².

The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of static friction µ is such that

80 N = µ (196 N) → µ = (80 N)/(196 N) ≈ 0.408

Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of kinetic friction ν is

40 N = ν (196 N) → ν = (40 N)/(196 N) ≈ 0.204

And so the closest answer is C.

(Note: µ and ν are the Greek letters mu and nu)

3 0

2 years ago

An object starts at the 100 m mark and ends up at the -100 m mark, 50

horsena [70]

It traveled 200 m in 50 seconds. 200/50 can be simplified to 4 m/s!
The velocity is -4 m/s (negative because it travelled from 100 to -100 or backwards)

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1 year ago

You have a series circuit powered by a 9V battery. If you double the amount of

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It gets 2 times weaker

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2 years ago

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