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3 years ago

20 POINTS WILL BE MARK BRAINLIEST ATTACHED BELOW GAS LAW WORKSHEET

Physics

1 answer:

german3 years ago

7 0

Explanation:

I will do two of each as examples.

Boyle's law says that at constant temperature, the product of the initial pressure and volume equals the product of the final pressure and volume.

1. P₁ V₁ = P₂ V₂

(1.5 atm) (10.0 L) = (0.75 atm) V

V = 20.0 L

2. P₁ V₁ = P₂ V₂

(100.0 kPa) (500.0 mL) = P (1,000.0 mL)

P = 50.0 kPa

Charles' law says that at constant pressure, the quotient of the initial volume and temperature equals the quotient of the final volume and temperature.

6. V₁ / T₁ = V₂ / T₂

(10.0 L) / (1500 K) = V / (750 K)

V = 5.0 L

7. V₁ / T₁ = V₂ / T₂

(500.0 mL) / (100 K) = (1000.0 mL) / T

T = 200 K

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A ball which is dropped from the top of a building strikes the ground with a speed of 30m/s. Assume air resistance can be ignore

Vika [28.1K]

Answer:

h= 45.87 m.

Explanation:

Data given:

Vf= 30m/s , and we know that g = 9.8 m/s²

The ball has an initial velocity of zero Vi = 0 m/s²

To Find:

Height of the building = ?

Solution:

According to 3rd law of the motion;

2aS= Vf² - Vi² ( S= h , a=g)

2*9.81*h = (30)² - (0)²

h= 45.87 m.

3 0

3 years ago

Which country sent the first artificial satellite into orbit?

Rasek [7]

the soviet union was the first country.

8 0

3 years ago

a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?

lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let $f$ be frictional force acting in the backward direction as shown. Let the coefficient of friction be $\mu$ . Let $N$ be the normal reaction force acting on the bag.

Given:

Mass of the bag is, $m=8.10\textrm{ kg}$

Force acting at $\theta = 38$ ° is $F= 29.5\textrm{ N}$

Acceleration due to gravity is, $g=9.8\textrm{ }m/s^{2}$

The force F can be resolved into its components as $F_{x}=F \cos \theta$ and $F_{y}=F \sin \theta$

Therefore,

$F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}$

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

$N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}$

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

$f=F_{x}\\f=23.25\textrm{ N}$

Now, frictional force is given as:

$f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38$

Therefore, the coefficient of friction is 0.38.

8 0

4 years ago

Un ladrón viaja en un auto a una velocidad excesiva de 30m/s, en dicho momento un policía de tránsito, lo observa y monta en su

tester [92]

Answer:

t = 12 s

Explanation:

To find the time in which the police reaches the thief you write the equation of motion of both thief and police:

The thief has a constant velocity, the position is then given by:

$x=vt$ (1)

The police has an acceleration, then the position is:

$x=v_ot+\frac{1}{2}at^2$ (2)

the time in which the police reaches the thief is when their positions are equal, that is, when expression (1) equals expression (2). But before you calculate the acceleration of the police:

$a=\frac{v-v_o}{t}\\\\v_o=0m/s\\\\a=\frac{50m/s}{10s}=5\frac{m}{s^2}$

you replace this values in (2) and you equal the expression (1) and (2) (with vo = 0):

$vt=\frac{1}{2}at^2\\\\\frac{1}{2}at^2-vt=0\\\\(\frac{1}{2}at-v)t=0$

one root is t = 0s, but it is omitted because is the momment in which the thief pass in front of the police. The other root is:

$\frac{1}{2}at-v=0\\\\t=\frac{2v}{a}=\frac{(2)(30m/s)}{5m/s^2}=12s$

hence, the time is 12 s

8 0

3 years ago

Why is their displacement the same when the distance each traveled was different?

motikmotik

Displacement is how far something is from its starting position

for example if X moved 10 feet, turned around and moved 30 feet, its distance would be 40, but the displacement would be 20,
and if Y Moved 2 feet, and turned around and moved 22 feet, its distance would be 24, but its displacement would also be 20

4 0

3 years ago