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muminat
3 years ago
13

Ice of mass 5 g at 0 °C melts to water at 0 °C.

Physics
1 answer:
amid [387]3 years ago
5 0

Answer:

Q=1670J

Explanation:

Mass of ice: m=5g=0.005kg

Latent heat: lambda=3.34×10⁵J/kg

Heat received by ice: Q=m×lambda

Q=0.005×3.34×10⁵=5×334=1670J

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Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax
kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its x-component is

F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its x-component is

F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N

So, the x-component of the resultant force is

F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its y-component is

F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its y-component is

F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N

So, the y-component of the resultant force is

F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

F=\sqrt{F_x^2+F_y^2}

Where in this problem, we have:

F_x=-7.14 N is the x-component

F_y=2.70 N is the y-component

And substituting, we find:

F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N

6 0
4 years ago
Read 2 more answers
What happens to the force of gravity between two masses if one mass is decreased?
Andrews [41]
<span>B. It stays the same</span>
7 0
3 years ago
Read 2 more answers
In a science fiction story, a microscopic black hole is given an enormous positive charge by firing an un-neutralized ion drive
Olegator [25]

Answer: distance d = 4.73e10m

Explanation: Suppose the charge on the black hole is 5740 C which is a positive charge.

Using electric potential V formula:

V = kq / d

Where K = 9.05×10^9Nm^2/C

And e = 1.6×10^-19C

But you don't need to substitute it.

1090 V = 8.99e9N·m²/C² * 5740C /d

Make d the subject of formula

d = 4.73e10 m

6 0
3 years ago
Which best describes most covalent compounds?<br> O soft<br> O brittle<br> 3<br> O cold<br> O warm
LUCKY_DIMON [66]

Answer: Brittle

Explanation:

took the test and I chose Soft, Soft is the wrong answer don't choose it. The CORRECT ANSWER IS BRITTLE

7 0
4 years ago
Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 tesla?
Nonamiya [84]

Complete Question:

When specially prepared Hydrogen atoms with their electrons in the 6f state are placed into a strong uniform magnetic field, the degenerate energy levels split into several levels. This is the so called normal Zeeman effect.

Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 Tesla?

Answer:

ΔE = 1.224 * 10⁻²² J

Explanation:

In the 6f state, the orbital quantum number, L = 3

The magnetic quantum number, m_{L} = -3, -2, -1, 0, 1, 2, 3

The change in energy due to Zeeman effect is given by:

\triangle E = m_{L} \mu_{B} B

Magnetic field B = 2.02 T

Bohr magnetron, \mu_{B} = 9.274 * 10^{-24} J/T

\triangle E = 6 * 9.274 * 10^{-24} * 2.2\\

ΔE = 1.224 * 10⁻²² J

5 0
3 years ago
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