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Ksivusya [100]
3 years ago
11

Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determi

ned by the equation d = 144 – 16t2, where t is the number of seconds it takes the car to travel down to each point on the ride. For which interval of time is Greg’s car moving in the air?
0 -3 t>3
0
Physics
1 answer:
Talja [164]3 years ago
3 0
The solution to the problem is as follows:

 <span>Average = 80 
So Sum = 80 * 5 = 400 
Mode = 88, so two results are 88 (if three results were 88, then the median would be 88). 
Three results are 81, 88, and 88. 
That leaves 143. We could still have one 81 score, so that leaves the lowest score as 62. 

Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determined by the equation d = 144 - 16t2, where t is the number of seconds it takes the car to travel down to each point on the ride. How many seconds will it take Greg to reach the ground? 
d = 144 - 16t2 
0 = 144 - 16t2 
16t^2=144 
t^2=9 
t=3</span>
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C.

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The correct answer is C that is . It flips from spin down to spin up.

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Because of changes over time, the most accurate weather forecasts are: A) analog forecasts. B) long-term forecasts. C) seven-day
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In some applications of ultrasound, such as its use on cranial tissues, large reflections from the surrounding bones can produce
worty [1.4K]

Answer:

b. 0.75 mm

Explanation:

The distance between antinodes d is half the wavelength \lambda. We can obtain the wavelength with the formula v=\lambda f, where f is the frequency given (f=1MHz=1\times10^6Hz) and v is the speed of sound in body tissues (v=1540m/s), so putting all together we have:

d=\frac{\lambda}{2}=\frac{v}{2f}=\frac{1540m/s}{2(1\times10^6Hz)}=0.00077m=0.77mm

which is very close to the 0.75mm option.

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4 years ago
There are five different types of muscles in exercise physiology.
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3 years ago
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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