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2 years ago

An advertiser who calls a basement apartment a “garden apartment” is using the rhetorical device called

1 answer:

3 0

I'd say B or D, depending on where the place is. If it's in South Kensington, London, then maybe D. If it's in a deprived area then I'd say B.

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If an object is moving eastward and slowing down, then the direction of its acceleration is

Alchen [17]

C. Eastward. Acceleration is the change in speed so it can be a positive (speeding up) or negative (slowing down) acceleration

4 0

2 years ago

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the

o-na [289]

The equation we use is mλ=dsinθ for intensity maximas. We are given at the first maximum (m=1), it occurs at 17.8 degrees. Thus we can solve for d by substituting known values into our equation.

(1) (632.8*10^-9m)=dsin(17.8) => d = 2.07*10^-6m

Next we want to find the angle at the second maximum (m=2) so we need to solve for θ.

(2) (632.8*10^-9m) = (2.07*10^-6m)sinθ

θ=37.69 degrees

Hopes this helps!

P.S. I hope this is right. If not sorry in advance.

3 0

3 years ago

Which one is correct and why?

Gelneren [198K]

'A' and 'C' are exactly the same circuit, except the voltmeter's terminals are flipped.

'A' is the correct way to hook everything up.

If you start at the positive terminal of the battery, and follow the flow of current through the circuit and around to the negative terminal, you're following the path where the voltage gets lower and lower and lower all the way.

So each time you come to any device in the circuit ... whether it's a resistor or a meter ... you would be hitting the positive side of it first, and then the voltage where you come out on the other side of it would be lower.

So the left side of the resistor is more positive, and the right side is more negative. The voltmeter is connected correctly in 'A', but it's backwards in 'C'. If you connect the voltmeter like in 'C' and turn things on, the voltmeter will try to go <em>down</em> from zero. You can't read the number on it, and It's possible that the voltmeter might be damaged.

6 0

3 years ago

A 0.5 kg mass moving East with a velocity of 8 m/s is acted upon by a force of 20 N for 2.0 sec.

Andrej [43]

Answer:b

Explanation:

7 0

2 years ago

A factory worker pushes a 32.0 kgkg crate a distance of 4.0 mm along a level floor at constant velocity by pushing horizontally

Luda [366]

Answer:

a) The force that the worker must apply has a magnitude of 75.317 newtons.

b) The external force does a work of 0.301 joules.

c) The friction force does a work of -0.301 joules.

d) Both normal force and gravity have done a work of 0 joules.

e) The total work done on the crate is 0 joules.

Explanation:

a) As the crate is moving at constant velocity, we know that magnitude of the force done on the crate must be equal to the friction force. Hence, we must use the following formula:

$F = \mu\cdot m\cdot g$ (1)

Where:

$F$ - External force, in newtons.

$\mu$ - Coefficient of kinetic friction, no unit.

$m$ - Mass, in kilograms.

$g$ - Gravity acceleration, in meters per square second.

If we know that $\mu = 0.24$ , $m= 32\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the external force is:

$F = (0.24)\cdot (32\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 75.317\,N$

The force that the worker must apply has a magnitude of 75.317 newtons.

b) The direction of the force is parallel to the direction of motion. The work done by this force ( $W_{F}$ ), in joules, is determined by this formula:

$W_{F} = F\cdot \Delta s$ (2)

Where $\Delta s$ is the distance travelled by the crate, in meters.

If we know that $F = 75.317\,N$ and $\Delta s = 0.004\,m$ , then the work done by the force is:

$W_{F} = (75.317\,N)\cdot (0.004\,m)$

$W_{F} = 0.301\,J$

The external force does a work of 0.301 joules.

c) The direction of the friction force is antiparallel to the direction of motion. The work done by this force ( $W_{f}$ ), in joules, is determined by this formula: