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Nat2105 [25]
3 years ago
10

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h

e have to get out of the way?
Physics
1 answer:
noname [10]3 years ago
7 0
The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 =  0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s
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At the surface, atmospheric pressure is 1.013 × 10^5 Pa. People can normally snorkel down to a depth of roughly one meter. What
natulia [17]

Answer:

1.01 × 10⁵ Pa  

Explanation:

At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.

We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.

Pressure at a depth is given by :

P=\rho gh

Where

\rho is the density of air, \rho=1.225\ kg/m^3

So,

P=1.225\times 9.8\times 1\\\\=12\ Pa

Total pressure, P = Atmospheric pressure + 12 Pa

= 1.013 × 10⁵ Pa + 12 Pa

= 1.01 × 10⁵ Pa

Hence, the total pressure is 1.01 × 10⁵ Pa.

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3 years ago
What is the period of a sound wave having a frequency of 340hz?
galben [10]
Period, T, is 1/frequency
T = 1/340 = 2.941ms
6 0
4 years ago
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CaHeK987 [17]
It is A or D but I believe A
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Suppose that you are taking a course in geometry. You are trying to determine whether two geometric shapes are identical, so you
Nataly_w [17]

Explanation:

Yes, in order to determine whether two geometric figures are identical or not we tend to rotate one of the figure clockwise or anti clock wise mentally. However, in clockwise rotation larger the angel more will be the time taken for instance, it will take longer to rotate 120° than to rotate 40°. Whereas in anti clockwise ration it will be vice versa.

3 0
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A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t
Arisa [49]

A) 2.4\cdot 10^{-16}kg

The radius of the oil droplet is half of its diameter:

r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m

Assuming the droplet is spherical, its volume is given by

V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3

The density of the droplet is

\rho=885 kg/m^3

Therefore, the mass of the droplet is equal to the product between volume and density:

m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg

B) 1.5\cdot 10^{-18}C

The potential difference across the electrodes is

V=17.8 V

and the distance between the plates is

d=11 mm=0.011 m

So the electric field between the electrodes is

E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

qE=mg

So, from this equation, we can find the charge of the droplet:

q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

q=-1.5\cdot 10^{-18}C

we can think this charge has made of N excess electrons, so the net charge is given by

q=Ne

where

e=-1.6\cdot 10^{-19}C is the charge of each electron

Re-arranging the equation for N, we find:

N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9

so, a surplus of 9 electrons.

3 0
3 years ago
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