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3 years ago

Which statement below accurately describes the mass of the solid in the diagram?

Physics

1 answer:

olasank [31]3 years ago

8 0

Answer:

The answer to your question is letter C. 243.8 g

Explanation:

We just need to add up to the weight of the different scales and subtract the value of the empty beaker.

The sum of the scales is: 200 + 60 + 8.8 = 268.8 g

Weight of the solid = total weight - empty beaker

Weight of the solid = 268.8 - 25 = 243.8 g

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A group of inexperienced weightlifters was randomized to either perform Crossfit or Olympic lifting workouts for 10 weeks. We th

lora16 [44]

The appropriate hypothesis test for this question is the two proportions test. As the name implies, it is often used to compare existing percentages between two groups. The raw data behind the percentages must be available as the sample size becomes crucially important in determining the test statistics.

The test statistic of the two-proportion test is the Z value. When the sample becomes very large, the Z value is governed under a normal function and is commonly known as the standardized Z value

5 0

3 years ago

A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.46 m, and is initially unc

KonstantinChe [14]

(a) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can solve the different part of the problem by using Gauss theorem.

Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:

$E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

where q is the charge contained in the spherical surface, so

$q=5.00 C$

Solving for E(r), we find the expression of the field for r<a:

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

(b) 0

The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.

So, if we use Gauss theorem for the region a < r < b, we get

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$

however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so

q' = + q - q = 0

And so we find

E(r) = 0

(c) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can use again Gauss theorem:

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$ (1)

where this time r > b (outside the shell), so the gaussian surface this time contained:

- the charge +q at the centre

- the inner surface, with a charge of -q

- the outer surface, with a charge of +q

So the net charge is

q' = +q -q +q = +q

And so solving (1) we find

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

which is identical to the expression of the field inside the shell.

(d) $-12.3 C/m^2$

We said that at r = a, a charge of -q is induced. The induced charge density will be

$\sigma_a = \frac{-q}{4\pi a^2}$

where $4 \pi a^2$ is the area of the inner surface of the shell. Substituting

q = 5.00 C

a = 0.18 m

We find the induced charge density:

$\sigma_a = \frac{-5.00 C}{4\pi (0.18 m)^2}=-12.3 C/m^2$

(e) $-1.9 C/m^2$

We said that at r = b, a charge of +q is induced. The induced charge density will be

$\sigma_b = \frac{+q}{4\pi b^2}$

where $4 \pi b^2$ is the area of the outer surface of the shell. Substituting

q = 5.00 C

b = 0.46 m

We find the induced charge density:

$\sigma_b = \frac{+5.00 C}{4\pi (0.46 m)^2}=-1.9 C/m^2$

3 0

3 years ago

Two objects are dropped from different heights at the same instant. If the first object takes 10.7 seconds to hit the ground and

Lubov Fominskaja [6]

Answer:

The answer to your question is : 521.8 m

Explanation:

Data:

Different heights

Time first object (tfo) = 10.7 s

Time second object (tso)= 14.8 s

Initial speed of both objects(vo) = 0 m/s

a = 9.81 m/s²

Formula:

h = vot + 1/2 (a)(t)² but vo = 0 so, h = 1/2 (a)(t)²

Then, height fo h = 1/2 (9.81)(10.7)² = 561.6 m

height so h = 1/2(9,81)(14.8)² = 1074.4 m

Difference in their heights = 1074.4 m - 561.6 m = 521.8 m

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3 years ago

A dam is a structure built across a river to hold back the river's water. The flow of water through a dam is controlled by gates

Bess [88]

Potential is the first blank and kinetic is the second blank.

This should be correct

8 0

3 years ago

Read 2 more answers

A home uses ten 100-watt lightbulbs for five hours per day. Approximately how many kilowatt-hours of electrical energy are consu

timurjin [86]

The electrical energy consumed in one year by using the light bulbs is 8,760 kWh.

The given parameters:

Number of light bulbs = 10
Power consumed by each bulb = 100 W
Time of energy consumption, t = 1 year

<h3>What is electrical energy?</h3>

This is the electric power consumed or dissipated at a given period of time.

The electrical energy consumed in one year by using the light bulbs is calculated as

E = Pt

$E = (100 \times 10) \times (1 \ yr \times \frac{8760 \ hrs}{1 \ yr} )= 8,760,000 \ W att-hours\\\\
E = \frac{8,760,000 }{1000} = 8,760 \ kWh$

Thus, the electrical energy consumed in one year by using the light bulbs is 8,760 kWh.

Learn more about electrical energy here: brainly.com/question/60890

4 0

2 years ago