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const2013 [10]
3 years ago
12

What is the area under the curve for the histogram below?

Physics
2 answers:
emmasim [6.3K]3 years ago
8 0

Answer:

it's 194

Explanation:

you have to add up every value of people through each hour of the day.

olga2289 [7]3 years ago
5 0
Well i honestly wouldn’t know
You might be interested in
ANSWER ASAP 50 POINTS
spin [16.1K]

Answer:It turns out the Venus flytrap is a power plant, capable of generating electrical signals. Each trap is actually a modified leaf: a hinged midrib, which would be the central vein of a more familiar leaf, joins the two lobes, which secrete a sweet sap to attract insects.

Explanation:The leaves of Venus' Flytrap open wide and on them are short, stiff hairs called trigger or sensitive hairs. When anything touches these hairs enough to bend them, the two lobes of the leaves snap shut trapping whatever is inside.

4 0
3 years ago
A boy can swim 3.0 meter a second in still water while trying to swim directly across a river from west to east, he is pulled by
lana66690 [7]

Answer:

Angle: 48.19^o

Explanation:

<u>Two-Dimension Motion</u>

When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.

Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.

To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

v_b\ cos\alpha

where v_b is the speed of the boy in still water and \alpha is the angle respect to the shoreline. If the river flows at speed v_s, we now set

v_b\ cos\alpha=v_s

\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}

\alpha=48.19^o

8 0
3 years ago
jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en
d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

\theta = \frac{2\pi}{6}

\theta = \frac{\pi}{3}

so the angular speed of the train is given as

\omega = \frac{\theta}{t}

\omega = \frac{\pi}{30} rad/s

now we have time period of the train given as

T = \frac{2\pi}{\omega}

T = \frac{2\pi}{\frac{\pi}{30}}

T = 60 s

5 0
3 years ago
Fl-19 in florida, if your pwc is equipped with an engine cut-off lanyard, what must you do with it?
melamori03 [73]

Answer: Attach it to clothing, personal flotation device or the person's person.

Explanation:

This particular Question or problem has to do with the rules and regulations or laws governing the use of Personal Watercraft(PWC) in the state of Florida in the United States of America. Hence, one of the rules that is applicable to the use of boats and waterways or the use of personal watercraft in Florida is that in florida, if ones pwc is equipped with an engine cut-off lanyard the person operating it must ATTACH IT TO THE CLOTHING, PERSONAL FLOATATION DEVICE IR THE OPERATORS' PERSON.

Another rule band the use of flammable personal flotation device with Personal Watercraft(PWC) in the state of Florida. All this rules and guildlines are made to limit or minimize hazards.

3 0
3 years ago
The current through a 0.2-H inductor is i(t) = 10te–5t A. What is the energy stored in the inductor?
lakkis [162]

Answer:

E = 10t^2e^-10t Joules

Explanation:

Given that the current through a 0.2-H inductor is i(t) = 10te–5t A.

The energy E stored in the inductor can be expressed as

E = 1/2Ll^2

Substitutes the inductor L and the current I into the formula

E = 1/2 × 0.2 × ( 10te^-5t )^2

E = 0.1 × 100t^2e^-10t

E = 10t^2e^-10t Joules

Therefore, the energy stored in the inductor is 10t^2e^-10t Joules

6 0
3 years ago
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