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aleksandrvk [35]
3 years ago
15

Which characteristics do temperate marine and temperate continental climates have in common? Check all that apply.

Physics
1 answer:
katrin2010 [14]3 years ago
6 0

Answer:

moderate amount of precipitation

two seasons with temperature differences

Explanation:

Temperate marine climates is also known as oceanic climate and areas who experience this climate are closer to oceans. There is relative precipitation all round the year due to the close proximity to the water body and have the hot and cold season.

The temperate continental climate also has two season and relative rainfall but found in the areas along the Northern hemispheres.

Both climates have moderate amount of precipitation and two seasons with temperature differences.

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In calisthenics, the resistance is ultimately provided by __________. A. free weights B. springs and elastic bands C. gravity D.
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Answer:

I think the answer is b am sorry if it is wrong

Explanation:

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Which form of energy does flowing water possess
Leni [432]

Answer:

hydro energy?

Explanation:

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A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the
insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}

Req =3.03Ω , R1 =12.18Ω

\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}

\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}

\frac{1}{R2}=0.2479

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

3 0
3 years ago
You are listening to the radio when one of your favorite songs comes on, so you turn up the volume. If you managed to increase t
andrew-mc [135]

To solve this problem we need to apply the corresponding sound intensity measured from the logarithmic scale. Since in the range of intensities that the human ear can detect without pain there are large differences in the number of figures used on a linear scale, it is usual to use a logarithmic scale. The unit most used in the logarithmic scale is the decibel yes described as

\beta_{dB} = 10log_{10} \frac{I}{I_0}

Where,

I = Acoustic intensity in linear scale

I_0 = Hearing threshold

The value in decibels is 17dB, then

17dB = 10log_{10} \frac{I}{I_0}

Using properties of logarithms we have,

\frac{17}{10} = log_{10} \frac{I}{I_0}

log_{10} \frac{I}{I_0} = 1.7

\frac{I}{I_0} = 10^{1.7}

\frac{I}{I_0} = 50.12 W/m^2

Therefore the factor that the intensity of the sound was 50.12W/m^2

5 0
3 years ago
A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how
Montano1993 [528]
<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity V_{o} has two components, because the brick was thrown at an angle \alpha=61\º:

V_{ox}=V_{o}cos\alpha   (1)

V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}  (2)

V_{oy}=V_{o}sin\alpha   (3)

V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}   (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

X=V_{ox}.t  (5)

Where:

X=10.1m is the distance where the brick landed

t is the time in seconds

If we already know X and V_{ox}, we have to find the time (we will need it for the following equation):

t= \frac{X}{ V_{ox}}  (6)

t=2.42s  (7)

<h2>In the y-axis: </h2>

-y=V_{oy}.t+\frac{1}{2}g.t^{2}   (8)

Where:

y is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

g=-9.8\frac{m}{s^{2}} is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}   (9)

-y=-10.52m   (10)

Multiplying by -1 each side of the equation:

y=10.52m >>>>This is the height of the building

3 0
3 years ago
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