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2 years ago

15

Jonathan answered 28 out of 35 questions correctly on his chemistry test. What is his percent score?

1 answer:

pav-90 [236]2 years ago

8 0

Seriously? It’s that hard to use a calculator??

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What type of forces accelerate masses?

leva [86]

Answer:

i would say that the answer would be B

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3 years ago

Find the angle between forces of 41 pounds and 68 pounds given a magnitude of 87 pounds for the resultant force. (Hint: Write fo

dedylja [7]

Answer:

$\theta = 76.9 degree$

Explanation:

As we know that the resultant of two vectors is given as

$R = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 cos\theta$

here we know that

$R = 87 Lb$

$F_1 = 41 Lb$

$F_2 = 68 Lb$

now we have

$87 = \sqrt{41^2 + 68^2 + 2(41)(68)cos\theta$

$87^2 = 6305 + 5576 cos\theta$

$\theta = cos^{-1}(\frac{1264}{5576})$

$\theta = 76.9 degree$

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2 years ago

How do spectroscopes help with studying of distance stars?

Svetradugi [14.3K]

Answer:

You take the light from a star, planet or galaxy and pass it through a spectroscope, which is a bit like a prism letting you split the light into its component colours. "It lets you see the chemicals being absorbed or emitted by the light source. From this you can work out all sorts of things," says Watson

8 0

2 years ago

At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spacesh

Svetradugi [14.3K]

Answer:

The time needed is $T = 16.8 s$

Explanation:

From the question we are told that

The magnitude of the stimulated acceleration due gravity is $a = 0.5 g$

The diameter of the spaceship is $d = 35m$

Generally the force acting on the spaceship is

$F = ma$

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

$F = \frac{mv^2}{r}$

Thus

$ma = \frac{mv^2}{r}$

=> $\frac{v^2}{r} = a$

Generally the speed of this spaceship is mathematically represented as

$v = \frac{2 \pi}{T}$

=> $v^2 = [\frac{2\pi}{T}] ^2$

=> $\frac{\frac{4\pi^2 r^2}{T^2} }{r} = 0.5g$

=> $\frac{4 \pi^2 r }{T^2} = 0.5 g$

=> $T = \sqrt{ \frac{4\pi^2 r}{0.5g}}$

substituting values

$T = \sqrt{ \frac{4* (3.142)^2 *(35)}{0.5 * 9.8}}$

$T = 16.8 s$

4 0

2 years ago

Read 2 more answers

A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be

lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × $10^{-3}$ m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ) ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v / 2 ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 ) / 2

0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 = 108

ω² = 108 / 7.160 × $10^{-5}$

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

7 0

3 years ago