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2 years ago

How does having waters of hydration affect the enthalpy for dissolving?

Physics

1 answer:

poizon [28]2 years ago

4 0

Answer:

it makes it so wet

Explanation:

hope it helps

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A straight wire segment 5 m long makes an angle of 30° with a uniform magnetic field of 0.37 T. Find the magnitude of the force

SIZIF [17.4K]

Answer:

The magnitude of the force on the wire is 2.68 N.

Explanation:

Given that,

Length of the wire, L = 5 m

Magnetic field, B = 0.37 T

Angle between wire and the magnetic field, $\theta=30^{\circ}$

Current in the wire, I = 2.9 A

We need to find the magnitude of the force on the wire. The magnetic force in the wire is given by :

$F=BIL\ \sin\theta\\\\F=0.37\ T\times 2.9\ A\times 5\ m\times \ \sin(30)\\\\F=2.68\ N$

So, the magnitude of the force on the wire is 2.68 N. Hence, this is the required solution.

7 0

3 years ago

What has more kinetic energy 15 kg ball rolling north at 15 m/s or a 15 kg ball rolling backwards at 7m/s

Setler79 [48]

Answer:

15 kg ball

Explanation:

6 0

3 years ago

F.r.e.e points :]]]

Colt1911 [192]

Ok? I don’t know what you want me to do though

4 0

2 years ago

Read 2 more answers

PLEASE HELP!!!!!!! A student is trying to demonstrate static electricity, so they rub two identical balloons with a neutral rabb

vlada-n [284]

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

4 0

2 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

5 0

3 years ago