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Liula [17]
3 years ago
5

A soda has a volume of 560 mL and a density of 3.2 g/mL. What is the mass?

Physics
1 answer:
kolbaska11 [484]3 years ago
8 0
The correct answer is: 1792g or 1800g.

(When you round it)
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Chinook salmon are able to move upstream faster by jumping out of the water periodically; this behavior is called porpoising. Su
deff fn [24]

Answer:he formula for average speed is (total distance/total time)

the y-component does not matter in this problem. so do 6.26(cos45)=4.43m/s to find the x-component velocity which is constant throughout the duration of the flight. the total distance is 2L because he travels distance L twice.

the total time is ((time in water)+(time out of water)) since you dont have time you must eliminate it. to do this you need (distance)/(time)=velocity

solve for time and you get T=D/V

time in water is L/3.52 and time out of water is L/4.43

add them together and you get (4.43L+3.52L)/(15.59) = 7.95L/15.59

that value is your total time

divide you total distance (2L) by total time (7.95L/15.59) and the Ls cancel out and you get

(31.18)/(7.95) = 3.92 m/s = Average Speed

Explanation:

7 0
3 years ago
Which is true about light and heat?
Snowcat [4.5K]

My best guess is c) Dark colors reflect less radiation making them warmer.

hope this helps!

5 0
3 years ago
Read 2 more answers
Does the zero electric field intensity in a given region imply zero potential?​
Rama09 [41]

Answer:

No, just because the electric field is zero at a particular point, it does not necessarily mean that the electric potential is zero at that point. ... At the midpoint between the charges, the electric field due to the charges is zero, but the electric potential due to the charges at that same point is non-zero.

Explanation:

7 0
3 years ago
What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =
iren2701 [21]
1) The electric potential at a distance r from a single point charge is given by
V(r) = k  \frac{q}{r}
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.

The charge in this problem is
q=2.2 \mu C =2.2 \cdot 10^{-6} C
So the potential at distance r=6.3 m is
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2 C^{-2})  \frac{2.2 \cdot 10^{-6} C}{6.3 m}=3139 V

2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2C^{-2})  \frac{2.2 \cdot 10^{-6} C}{99 m}=198 V
3 0
3 years ago
Read 2 more answers
A kg object has 400 j of potential energy. Find how high off the groumd the object is
blondinia [14]
Potential energy = mgh
Energy, U = 400J
g = 10m/s^2
m = A kg
h = U ÷ mg
= 400÷10A
= 40÷A

mass is not given
6 0
3 years ago
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