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sladkih [1.3K]
3 years ago
15

A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0 m high cliff. At the instant the ball i

s thrown, a woman starts running away from the base of the cliff with a constant speed of 6.00 m/s. The woman runs in a straight line on level ground, and air resistance acting on the ball can be ignored.How far does the woman run before she catches the ball?
Physics
1 answer:
scoray [572]3 years ago
3 0

Answer:

The distance traveled by the woman is 34.1m

Explanation:

Given

The initial height of the cliff

yo = 45m final, positition y = 0m bottom of the cliff

y = yo + ut -1/2gt²

u = 20.0m/s initial speed

g = 9.80m/s²

0 = 45.0 + 20×t –1/2×9.8×t²

0 = 45 +20t –4.9t²

Solving quadratically or by using a calculator,

t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s

So this is the total time it takes for the ball to reach the ground from the height it was thrown.

The distance traveled by the woman is

s = vt

Given the speed of the woman v = 6.00m/s

Therefore

s = 6.00×5.69 = 34.14m

Approximately 34.1m to 3 significant figures.

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Elemental analysis of the unknown gas from part a revealed that it is 30.45% n and 69.55% o by mass. What is the molecular formu
Anna71 [15]

Answer:

the molecular formula for the gas is NO₂

Explanation:

since it contains

Nitrogen = n → 30.45%

Oxygen = o → 69.55%

and 30.45%+69.55% = 100% , then the gas only contains nitrogen and oxygen

Also we know that the proportion of oxygen over nitrogen  is

proportion of oxygen over nitrogen  = moles of oxygen / moles of nitrogen

since

moles = mass / molecular weight

then for a sample of 100 gr of the unknown gas

mass of oxygen = 69.55%*100 gr = 69.55 gr

mass of Nitrogen = 30.45%*100 gr = 30.45 gr

proportion of oxygen over nitrogen = (mass of oxygen/ molecular weight)/(mass of nitrogen / molecular weight of nitrogen ) =  (69.55 gr/ 16 gr/mol) /( 30.45 gr /14 gr/mol) = 1.998 mol of O/ mol of N≈ 2 mol of O/ mol of N

therefore there are 2 atoms of oxygen per atom of nitrogen

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6 0
3 years ago
If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can st
liberstina [14]

Answer:

The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Explanation:

Given;

coefficient of kinetic friction, μ = 0.84

speed of the automobile, u = 29.0 m/s

To determine the  the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration

x is the shortest distance

First we determine a;

From Newton's second law of motion

∑F = ma

F is the kinetic friction that opposes the motion of the car

-Fk = ma

but, -Fk = -μN

-μN = ma

-μmg = ma

-μg = a

- 0.8 x 9.8 = a

-7.84 m/s² = a

Now, substitute in the value of a in the equation above

v² = u² + 2ax

when the automobile stops, the final velocity, v = 0

0 = 29² + 2(-7.84)x

0 = 841 - 15.68x

15.68x = 841

x = 841 / 15.68

x = 53.64 m

Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

4 0
2 years ago
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