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3 years ago

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What will a spring scale read for the weight of a 57.0-kg woman in an elevator that moves upward with constant speed of 5.0 m/s

?

1 answer:

Tanzania [10]3 years ago

4 0

The spring scale will read 559 Newton's or 125.7 pounds.

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How do I count atoms in science

VashaNatasha [74]

Answer:

The "2" tells us that there are 2 hydrogen atoms in this compound.

Explanation:

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2 years ago

If your friend drops a chocolate bar to you from a height of 5.0 m above your hands,

Sladkaya [172]

Answer:

<h3>1.01 s</h3>

Explanation:

Using the equation of motion S = ut+1/2gt² to solve the problem where;

u is the initial velocity of the chocolate = 0m/s

t is the time taken

g is the acceleration due to gravity = 9.81m/s²

S is the height of fall = 5.0m

Substituting the given parameter into the formula to get the time t we have;

5 = 0(t)+1/2(9.81)t²

5 = 4.905t²

t² = 5/4.905

t² = 1.019

t = √1.019

t = 1.009 secs

<em>Hence it will take 1.01 secs for me to catch the chocolate bar</em>

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3 years ago

A typical oil control ring consists of _______ separate part(s).

azamat

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5 0

2 years ago

Read 2 more answers

A car is running at the speed of 45km/hr see a child 25 meter ahead and suddenly apllies a brakes. If the retradation of the car

kenny6666 [7]

Answer:

The car stops in 7.78s and does not spare the child.

Explanation:

In order to know if the car stops before the distance to the child, you take into account the following equation:

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

vo: initial speed of the car = 45km/h

a: deceleration of the car = 2 m/s^2

t: time

xo: initial distance to the child = 25m

x: final distance to the child = 0m

It is necessary that the solution of the equation (1) for time t are real.

You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:

$45\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=12.5\frac{m}{s}$

$0=25+12.5t-2t^2\\\\2t^2-12.5t-25=0\\\\t_{1,2}=\frac{-(-12.5)\pm \sqrt{(-12.5)^2-4(2)(-25)}}{2(2)}\\\\t_{1,2}=\frac{12.25\pm 18.87}{4}\\\\t_1=7.78s\\\\t_2=-1.65s$

You take the first value t1 because it has physical meaning.

The solution for t is real, then, the car stops in 7.78s and does not spare the child.

4 0

3 years ago

A 1200-kg car initially at rest undergoes constant acceleration for 8.8 s, reaching a speed of 10 m/ s. It then collides with a

atroni [7]

Answer:

The final kinetic energy of the two-car system is 60,000 J.

Explanation:

Given;

mass of the car, m = 1200 kg

time of motion, t = 8.8 s

final velocity of the car, v = 10 m/s

Apply the principle of conservation of kinetic energy; the initial kinetic energy is equal final kinetic energy.

$K.E_i = K.E_f\\\\K.E_f = \frac{1}{2}mv^2\\\\K.E_f = \frac{1}{2}(1200)(10)^2\\\\K.E_f = 60,000 \ J$

Therefore, the final kinetic energy of the two-car system is 60,000 J.

4 0

2 years ago