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OlgaM077 [116]
3 years ago
11

A 216-m-wide river flows due east at a uniform speed of 2.2 m/s. A boat with speed of 7.9 m/s relative to the water leaves the s

outh bank of the river pointed in a direction 34 degrees west of north. What is the magnitude of the boat's velocity relative to the ground (in m/s)
Physics
1 answer:
Mademuasel [1]3 years ago
7 0

Answer:

The magnitude of the velocity of boat w.r.t ground is 6.2015 m/s

Solution:

As per the question;

Width of the river, w = 216 m

Uniform speed of boat due east, u_{x} = 2.2\hat{i} m/s

Angle in north west direction, \theta = 34^{\circ}

Velocity of the boat relative to water, v_{bw} = 7.9m/s

Now,

Velocity of the boat relative to the water is given by:

v_{bw} = -7.9cos34^{\circ}\hat{i} + 7.9sin34^{\circ}\hat{j}

v_{bw} = - 6.55\hat{i} + 4.42\hat{j} m/s

Also, velocity of water w.r.t ground is v_{wg} = u_{x} = 2.2\hat{i} m/s

Now,

Magnitude of the velocity of boat w.r.t ground is given by:

v_{bg} = v_{bw} + v_{wg}

v_{bg} = - 6.55\hat{i} + 4.42\hat{j} + 2.2\hat{i}

v_{bg} = - 4.35\hat{i} + 4.42\hat{j} m/s

|v_{bg}| = \sqrt{(- 4.35)^{2} + (4.42)^{2}} = 6.2015\ m/s

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3 years ago
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a. How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter t
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Explanation:

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3 years ago
NEED HELP !!!!!
goblinko [34]

Answers with Explanations:

1. Answer: True

Explanation:

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2. Answer: True

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Applying pressure also <em>decreases swelling </em>because it restricts the blood flow and fluid from reaching into the injury.

3. Answer: False

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4. Answer: False

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5. Answer: True

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6. Answer: False

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7. Answer: False

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9. Answer: True

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10. Answer: True

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11. Answer: all of the above

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12. Answer: True

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13. Answer: False

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14. Answer: True

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15. Answer: True

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16. Answer: True

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17. Answer: False

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4 0
3 years ago
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Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = x+y=149.6\times 10^6\ m

G = Gravitational constant

M_s = Mass of Sun = 1.989\times 10^{30}

M_e = Mass of Earth = 5.972\times 10^{24}\ kg

According to the question

\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y

x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m

The probe should be 258774.9441 m from Earth

7 0
3 years ago
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