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kaheart [24]
2 years ago
14

Jim stands beside a wide river and wonders how wide it is. he spots a large rock on the bank directly across from him. he then w

alks upstream 65 strides and judges that the angle between him and the rock, which he can still see, is now at an angle of 30o . jim measures his stride to be about 0.8 m long. estimate the width of the river.
Physics
1 answer:
LuckyWell [14K]2 years ago
7 0

To solve this problem, we must imagine that Jim’s initial position, the position of the rock, and Jim’s final position all connects to form a triangle. Now we can imagine that the triangle is a right triangle with the 90° angle on the initial position.

The angle of 30° is directly opposite to the length of his total stride while the width of the river is the side adjacent to the angle. Therefore can use the tan function to solve for the width of the river:

tan θ = opposite side / adjacent side

tan 30 = total stride distance / width of river

where total stride distance = 65 * 0.8 = 52 m

width of river = 52 m / tan 30

<span>width of river = 90.07 m</span>

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A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a
UNO [17]

Answer:

P_{avg} = 6.283*10^{-9} \ W

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})

where;

E= \frac{d \phi}{dt}

E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

P_{avg} = 6.283*10^{-9} \ W

6 0
2 years ago
Some gliders are launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. Wha
aliina [53]

Answer:

P=627.47W

Explanation:

To solve this problem we have to take into account, that the work done by the winch is

W=Fh

the force, at least must equal the gravitational force

F=Mg=(156kg)(9.8\frac{m}{s^2})=1258.8N

with force the tension in the cable makes the winch go up.

The work done is

W=(1258.8N)(58.0m)=73010.4J

To calculate the power we need to know what is the time t. But first we have to compute the acceleration

The acceleration will be

v_f^2=v_0+2ah\\a=\frac{v_f^2}{2h}=\frac{(24.9\frac{m}{s})}{2(58.0m)}=0.214\frac{m}{s^2}

and the time t

v_f=v_0+at\\t=\frac{v_f}{a}=116.35s

The power will be

P=\frac{W}{t}=\frac{73010.4J}{116.35s}=627.47W

HOPE THIS HELPS!!

6 0
2 years ago
Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet
sveta [45]

Answer:

The level of the root beer is dropping at a rate of 0.08603 cm/s.

Explanation:

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r is the radius of the cone

h is the height of the cone

The ratio of the radius and the height remains constant in overall the cone.

Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm

h = 13 cm

r / h = 5 / 13

r = {5 / 13} h

V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h

V=\frac {550}{3549}\times h^3

Also differentiating the expression of volume w.r.t. time as:

\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}

Given: \frac {dV}{dt} = -4 cm³/sec (negative sign to show leaving)

h = 10 cm

So,

-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}

\frac{55000}{1183}\times \frac {dh}{dt}=-4

\frac {dh}{dt}=-0.08603\ cm/s

<u>The level of the root beer is dropping at a rate of 0.08603 cm/s.</u>

3 0
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Licemer1 [7]

Answer:

24

Explanation:

4 0
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Ignoring fluid resistance, football will <span>maintain a constant speed until other forces accelerate the football.</span>
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2 years ago
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