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Natali [406]
2 years ago
10

Which of the following are scalar quantities? select all that apply

Physics
1 answer:
dolphi86 [110]2 years ago
3 0

Answer:

1, 4, 5, see the explanation below

Explanation:

We must remember that scalar magnitudes are distinguished by having only a physical quantity, that is, they have no sense or direction as an example of scalar quantities, we find mass, temperature, energy, specific heat, power among others.

1 . 150 [grams] , because is a mass =  scalar

4. 5 kilometer [race], is an amount = scalar

5. 34 steps, is an amount = scalar

Number 2, and 3 are vectors because they have amount and direction.

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When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical powe
ch4aika [34]

Answer:

4.41 W

Explanation:

P = IV, V = IR

P = V² / R

Given that P = 0.0625 when V = 1.50:

0.0625 = (1.50)² / R

R = 36

So the resistor is 36Ω.

When the voltage is 12.6, the power consumption is:

P = (12.6)² / 36

P = 4.41

So the power consumption is 4.41 W.

5 0
2 years ago
Learning Goal: How do 2 ordinary waves build up a "standing" wave? A very generic formula for a traveling wave is: y1(x,t)=Asin(
zheka24 [161]

Answer:

Explanation:

=Asin(kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!

Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.

Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From

Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the strJHJMNMMUJJHTGGHing?

a.There is no energy stored in the string: The string will remain straight for all subsequent times.

b.Energy will flow into the string, causing the standing wave to form at a later time.

c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.

d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.

3 0
2 years ago
If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporizat
Zanzabum

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

MM = 18 g/mol

The enthalpy in mass:

ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g

Finally, solving for m:

m = 970 / 2.261 = 429 g

Converting this into volume:

429 g = 429 mL

429 / 1000 = 0.429 L of water

3 0
2 years ago
A race car driver must average 200km/hr for four laps to qualify for a race. Because of engine trouble, the car averages only 17
vampirchik [111]
The average speed would have to be 260 km/hr due to the driver originally going 30 km/hr too slow the first two laps
5 0
2 years ago
Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi
Tju [1.3M]

By law of refraction we know that image position and object positions are related to each other by following relation

\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}

here we know that

\mu_1 = 1.473

h_o = 5 cm

\mu_2 = 1

now by above formula

\frac{1.473}{5} = \frac{1}{h_i}

h_i = 3.39 cm

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

7 0
2 years ago
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