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2 years ago

A 20 kg mass is dropped from a tall rooftop and accelerates at 9.8 m/s2. What is the weight of the dropped object?

2 answers:

7 0

It would have no “weight” since weight is determined by the force normal and since it is accelerating there is no normal force making it weightless

patriot [66]2 years ago

6 0

Objects in free fall are weightless.

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What is the difference between accurate data and reproducible data?

vladimir1956 [14]

Accuracy is a general concept while precision is more of a mathematical concept.

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2 years ago

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What is the IMA of the following pulley system?<br><br>34567

Lynna [10]

Answer:

IMA of given system = $\frac{F_{r} }{F_{e} }$

Explanation:

The "Ideal Mechanical advantage" (IMA) of given pulley is $\frac{F_{r} }{F_{e} }$ .

Ideal Mechanical advantage of a system is defined by the ratio of achieved or output force to the implied force. In the pulley system above, output force is the resistant force denoted by $F_{r}$ . The input force is analogous or equivalent to the effort applied i.e. $F_{e}$ .

Hence by dividing these two forces we calculate the IMA of the above mentioned pulley system which is $\frac{F_{r} }{F_{e} }$ .
Its mathematical reference would be:

IMA = $\frac{F_{r} }{F_{e} }$

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3 years ago

A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre

Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

$h=\sqrt{(36m)^2+(227m)^2}=229.83m$ (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

$y=y_o+v_ot+\frac{1}{2}gt^2$ (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

$3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s$

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

$v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}$

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

$\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°$

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0

3 years ago

A nasa spacecraft measures the rate r of at which atmospheric pressure on mars decreases with altitude. the result at a certain

Lesechka [4]

Answer: $4.21 \times 10^{-10} J/cm^4$

$1 kPa= 10^3 Pa$

$1 km=10^5 cm$

$1kPa/km=0.01 Pa/cm$

$1kPa/km=10^{-8} J/cm^4$

$\Rightarrow r= 0.0421 kPa/km= 0.0421 kPa/km \times \frac{10^{-8} J/cm^4}{1 kPa/km}= 0.0421 \times 10^{-8}J/cm^4=4.21 \times 10^{-10} J/cm^4$

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3 years ago

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A roller coaster is stationary at the top of a ramp.

Serjik [45]

Answer:

Explanation:

The roller coaster is stationary so the kinetic energy would be zero, but it is at the top of ramp so the potential energy would be high as its gravitational so it would have to be A

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2 years ago