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Mrac [35]
3 years ago
13

A boat crosses a river of width 161 m in which the current has a uniform speed of 1.09 m/s. The pilot maintains a bearing (i.e.,

the direction in which the boat points) perpendicular to the river and a throttle setting to give a constant speed of 2.77 m/s relative to the water. What is the magnitude of the speed of the boat relative to a stationary shore observer? Answer in units of m/s.

Physics
1 answer:
hoa [83]3 years ago
7 0

Answer:

|v|=2.98m/s

Explanation:

In order to solve this problem we must first draw a diagram that will represent the situation. (See picture attached)

In this case the boat is being dragged by the river while it has a given velocity relative to the water and perpendicular to the flow of the river. We can take these velocities as vectors so we can add them up:

|v|=\sqrt{(1.09m/s)^{2}+(2.77m/s)^{2}}

which yields:

|v|=2.98m/s

so an observer that is stationary on the shore will se the boat having a speed of 2.98m/s.

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A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.79 rad/s . Its total moment of inertia is 1790
Nezavi [6.7K]

Answer:

w_2=0.467rad/s

Explanation:

Four people standing on the ground each of mass and usually this questions have to find the final angular velocity

m_t=4*70kg=280kg

The radius r=4.2/2=2.1m

Angular velocity  w_1=0.79rad/s

The moment of inertia total is I_t=1790 kg/m^2

Momento if inertia

I_1=m_t*r^2

I_1=280kg*(2.1m)^2=1234.8kg*m^2

Angular momentum

I_1*w_1=I_t*w_2

Solve to w2

w_2=\frac{I_1*w_1}{I_t}

w_2=\frac{1790kg*m^2*0.79rad/s}{3024.8kg*m^2}

w_2=0.467rad/s

8 0
3 years ago
Total internal reflection will occur when:
ollegr [7]

Answer:

Try B or C if I'm wrong sorry

Explanation:

3 0
2 years ago
If the mass of the spacecraft were doubled, how much would the force of gravity would change?
vodka [1.7K]
F=mg
if m doubled, F would double as well

5 0
3 years ago
A disk-shaped merry-go-round of radius 2.13 m and mass 175 kg rotates freely with an angular speed of 0.651 rev/s. A 55.4 kg per
GREYUIT [131]

Answer:

Explanation:

To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum

I₁ ω₁ + I₂ ω₂ = ( I₁  + I₂ ) ω

I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .

I₁ = 1/2 mr²

= .5 x 175 x 2.13²

= 396.97 kgm²

I₂ = m r²

= 55.4 x 2.13²

= 251.34 mgm²

ω₁ = .651 rev /s

= .651 x 2π rad /s

ω₂ = tangential velocity of man / radius of disc

= 3.51 / 2.13

= 1.65 rad/s

I₁ ω₁ + I₂ ω₂ = ( I₁  + I₂ ) ω

396.97 x  .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω

ω = 3.14 rad /s

kinetic energy = 1/2 I ω²

= 3196 J

8 0
4 years ago
Explain how balanced and unbalanced forces effect an objects motion differently
Kazeer [188]
Balanced Forces acting on an object will not change the object's motion. Unbalanced Forces acting on an object will change the change the object's motion.
5 0
3 years ago
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