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djyliett [7]
3 years ago
10

Which value is equivalent to 7.2 kilograms? A. 720 grams B. 7,200 grams C. 72,000 grams D. 720 milligrams E. 7,200 milligrams

Physics
2 answers:
Aneli [31]3 years ago
8 0
1 kilogram = 1,000 grams
1 kilogram = 1,000,000 milligrams
Setler79 [48]3 years ago
6 0

If 1 kilogram is = to 1000 grams then 7.2 would = 7,200 because 7.2*1000=7,200 so your answer would be B

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Alla [95]

Answer:

T = 29.6 N

Explanation:

length of the rope is

L = 18 m

mass of the rope is

m = 12 kg

now we have

mass per unit length of the rope is given as

[te]\lambda = \frac{12 kg}{18 m}[/tex]

now time taken by wave to reach from end to other

t = \frac{L}{v}

2.7 s = \frac{18}{v}

v = 6.67 m/s

now we have

v = \sqrt{\frac{T}{\lambda}}

6.67 = \sqrt{\frac{T}{0.67}}

so we will have

T = 29.6 N

5 0
4 years ago
At the instant a traffic light turns green, a car starts with a constant acceleration of 1.3 m/s2. At the same instant a truck,
Bas_tet [7]

Answer:

(a) Distance traveled = 75.3846 m

(b) Velocity of car at that instant will be 14 m/sec

Explanation:

We have given acceleration of the car a=1.3m/sec^2

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(a) According to second equation of motion we know that s=ut+\frac{1}{2}at^2

So distance traveled by car s_c=0\times t+\frac{1}{2}\times 1.3t^2=0.65t^2

As the truck is moving with constant speed

So distance traveled by truck s_t=ut=7t

As the truck overtakes the car

So s_c=s_t

0.65t^2=7t

t=10.769sec

So distance traveled s_c=s_t=7\times 10.769=75.3846m

(b) From second equation of motion we know that v = u+at

So v = 0+1.3×10.769 = 14 m /sec

7 0
3 years ago
If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa
shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

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So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0
3 years ago
What is the molar mass of an ideal gas if a 0.800 g sample of this gas occupies a volume of 200. mL at 50.0 oC and 720. mm Hg?
Paladinen [302]
PV=nRT
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(0.1894736842)=(0.800/x)(0.08206)(323.15)
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7 0
3 years ago
Which wave characteristic determines color?
Veseljchak [2.6K]
I would said A is the best option if i’m wrong sorry
3 0
3 years ago
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