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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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A positive acceleration is defined as change of velocity upwards the direction that the object has taken. Therefore, a car having a positive acceleration means that the car is speeding up in the same direction as it traveling.

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If the force of gravity suddenly stopped acting on planets, they would
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Which of these belongs to a tropic level with the most energy?
MrRa [10]

Answer:

Berries is the correct answer because it is the produce in your pyramid and as each living thing is devoured by another there is less energy. For instance the berry has the most energy because it’s energy has just come from the sun. But then an insect eats it and consumes most of its energy but some energy is released into the atmosphere. Then a rodent eats the bug and consumes its energy but yet again some energy is released into the atmosphere. So each time there is less and less energy. Does that help any?

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8 0
3 years ago
determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund
wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is a_G= 6.78 m/s^2

<u>Explanation</u>:

N_A+N_B-Mg = 0

-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0

0.8N_B +0.8N_A = 975a_G

N_A+N_B = 9564.75 -------------(1)

-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0

-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0

-2.26N_A -0.06N_B= 0 ----------------(2)

Solving the equation (1) and(2)

N_A + N_B = 9564.75

-2.26N_A-0.06N_B=0

N_A = -260.85N

N_B = 9825.60N

\mu_s N_B + \mu_s N_A = 975a_G

0.8(9825.60)+0.8(-260.85) = 975a_Ga_G=\frac{7651.8}{975}a_G_1=7.4848m/s^2

Next lets assume that the front wheels contact with the ground N_A = 0

F_B = Ma_G

N_B = M_g

N_B - M_g = 0

N_B(b-a) –F_Bh = 0

F_B = 975a_G

N_B-975(9.8) = 0

N_B=9564.75N

9564.75(2.20 -1.82) -F_B(0.55)=0

\frac{3634.605}{0.55}=F_B

F_B = 6608.3

F_B = Ma_G

6608.3 = 975a_G

a_G = 6.7778 m/s^2

a_G_2 = 6.78m/s^2

Choosing the critical case

a_G = min(a_G_1 ,a_G_2)

a_G = min(7.848, 6.78)

a_G= 6.78 m/s^2

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