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3 years ago

How are Earth, the Moon and Sun alligned during a neap tide?

1 answer:

6 0

Neap tides occur at or immediately after the First Quarter
and Third Quarter moon phases. At those times, the moon
is 'to the side' of the Earth. I mean, the sun, Earth, and Moon
make a right angle, with Earth at the vertex.

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Help me with the question b.

Morgarella [4.7K]

Answer:

a) The specific heat capacity means the amount of heat needed by a unit mass of a material to increase its temperature in one unit.

b) Liquid P - $Q = 3840\,J$ , Liquid Q - $Q = 5500\,J$ , Liquid R - $Q = 7800\,J$ , Liquid S - $Q = 2856\,J$

Explanation:

a) The specific heat capacity means the amount of heat needed by a unit mass of a material to increase its temperature in one unit.

b) Let suppose that heat transfer rates between liquids and surroundings are stable. The quantity of the heat released is determined by the following expression:

$Q = m\cdot c\cdot (T_{r} - T_{f})$ (1)

Where:

$m$ - Mass of the liquid, in kilograms.

$c$ - Specific heat capacity, in joules per kilogram-degree Celsius.

$T_{r}$ - Initial temperature of the sample, in degrees Celsius.

$T_{f}$ - Freezing point, in degrees Celsius.

Liquid P ( $m = 1\,kg$ , $c = 160\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{r} = 30\,^{\circ}C$ , $T_{f} = 6\,^{\circ}C$ )

$Q = (1\,kg)\cdot \left(160\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 6\,^{\circ}C)$

$Q = 3840\,J$

Liquid Q ( $m = 1\,kg$ , $c = 220\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{r} = 30\,^{\circ}C$ , $T_{f} = 5\,^{\circ}C$ )

$Q = (1\,kg)\cdot \left(220\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 5\,^{\circ}C)$

$Q = 5500\,J$

Liquid R ( $m = 1\,kg$ , $c = 300\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{r} = 30\,^{\circ}C$ , $T_{f} = 4\,^{\circ}C$ )

$Q = (1\,kg)\cdot \left(300\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 4\,^{\circ}C)$

$Q = 7800\,J$

Liquid S ( $m = 1\,kg$ , $c = 102\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{r} = 30\,^{\circ}C$ , $T_{f} = 2\,^{\circ}C$ )

$Q = (1\,kg)\cdot \left(102\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 2\,^{\circ}C)$

$Q = 2856\,J$

6 0

2 years ago

A gas of helium atoms at 273 k is in a cubical container with 25.0 cm on a side. (a) what is the minimum uncertainty in momentum

qwelly [4]

wave function of a particle with mass m is given by ψ(x)={ Acosαx −

2α

≤x≤+

2α

0 otherwise , where α=1.00×1010/m.

(a) Find the normalization constant.

(b) Find the probability that the particle can be found on the interval 0≤x≤0.5×10−10m.

(d) Find its average momentum.

(e) Find its average kinetic energy −0.5×10−10m≤x≤+0.5×10−10m.

6 0

2 years ago

V02 Zone 90-100% MHR; =BPM

shepuryov [24]

Don’t use those links!!! there scams

3 0

2 years ago

1. An object in free fall will have an initial velocity equal to zero when: a. It is thrown vertically down

Ne4ueva [31]

Answer:

b. It is dropped

Explanation:

If the initial velocity is zero, the object move from rest. That happens if the object is dropped

6 0

2 years ago

A 25-kg child sits at the top of a 4-meter slide. After sliding down, the child is traveling at 5 m/s. How much PE does he start

Semmy [17]

Daniddmelo says it right there, don't know why he got reported.

The potential energy (PE) is mass x height x gravity. So it would be 25 kg x 4 m x 9.8 = 980 joules. The child starts out with 980 joules of potential energy. The kinetic energy (KE) is (1/2) x mass x velocity squared. KE = (1/2) x 25 kg x 5 m/s2 = 312.5 joules. So he ends with 312.5 joules of kinetic energy. The Energy lost to friction = PE - KE. 980- 312.5 = 667.5 joules of energy lost to friction.

Please don't just copy and paste, and thank you Dan cause you practically did it I just... elaborated more? I dunno.

4 0

2 years ago