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sukhopar [10]
3 years ago
12

A small mailbag is released from a helicopter that is descending steadily at 2.52 m/s. (a) After 4.00 s, what is the speed of th

e mailbag
Physics
1 answer:
gogolik [260]3 years ago
6 0

Answer: 41.72m/s

Explanation: v= u + gt

V = 2.52 + 9.8(4.00)

V = 41.72m/s

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A 60-kg cheetah reaches a speed of 30 m/s as it chases its prey.What is the kinetic energy of the cheetah?
Rama09 [41]
The answer would be 27,000 Joules because (1/2) m v^2 =30*900 which equals 27,000 J
4 0
3 years ago
Read 2 more answers
Compare and contrast the theories about the origin of the universe
Vaselesa [24]

Answer:

The best-supported theory of our universe's origin centers on an event known as the big bang. This theory was born of the observation that other galaxies are moving away from our own at great speed in all directions, as if they had all been propelled by an ancient explosive force.

Explanation:

hope this helps tho i don't quite know what you mean

7 0
3 years ago
The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is
stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

\rho_m = 846lb/ft^3

g = 32.17405ft/s^2

h_1 = 1in = \frac{1}{12} ft

For the air the defined properties would be

\rho_a = 0.0075lb/ft^3

g = 32.17405ft/s^2

h_2 = ?

We have for equilibrium that

\text{Pressure change in Air}=\text{Pressure change in Mercury}

\rho_m g h_1 = \rho_a g h_2

Replacing,

(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)

Rearranging to find h_2

h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}

h = 9400ft

Therefore the elevation of the mountain top is 9400ft

7 0
3 years ago
Assume that the function x(t) represents the length of tape that has unwound as a function of time. find θ(t), the angle through
bekas [8.4K]
We know that arc length (x(t)) is given with the following formula:
x(t)=\theta(t) r
Where r is the radius of the barrel. We must keep in mind that as barrel rolls its radius decreases because less and less tape is left on it.
If we say that the thickness of the tape is D then with every full circle our radius shrinks by d. We can write this down mathematically:
r(\theta)=r_0-\frac{D\cdot \theta}{2\pi}
When we plug this back into the first equation we get:
x(t)=\theta(r_0-\frac{D\theta}{2\pi}})\\ \frac{D\theta^2}{2\pi}-\theta r_0+x(t)=0\\
We must solve this quadratic equation.
The final solution is:
\theta=\frac{\pi r_0+\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D},\:\theta=\frac{\pi r_0-\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D}
It is rather complicated solution. If we asume that the tape has no thickness we get simply:
x(t)=\theta(r_0-\frac{D\theta}{2\pi}});D=0\\
x(t)=\theta r_0\\
\theta(t)=\frac{x(t)}{r_0}

8 0
4 years ago
Among all the electromagnetic waves (EM), which has the highest frequency? *
puteri [66]

first is gamma Ray's, last is d

8 0
3 years ago
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