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alexira [117]
3 years ago
12

Suppose an object is moving 2.1 m/s north on a river, but the river is flowing to the east at a velocity of 1.2 m/s. What is the

magnitude of the resultant velocity?
Physics
1 answer:
rewona [7]3 years ago
8 0

Answer:

2.4 m/s

Explanation:

Given:

Velocity of the object moving north = 2.1 m/s

Velocity of the river moving eastward = 1.2 m/s

The resultant velocity is the vector sum of the velocities of object and river.

Since the directions of velocity of object and river are perpendicular to each other, the magnitude of the resultant velocity is obtained using Pythagoras Theorem.

The velocities are the legs of the right angled triangle and the resultant velocity is the hypotenuse.

The magnitude of the resultant velocity (R) is given as:

R^2=2.1^2+1.2^2\\\\R^2=4.41+1.44\\\\R=\sqrt{5.85}\\\\R=2.4\ m/s

Therefore, the resultant velocity has a magnitude of 2.4 m/s.

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A soccer ball is sitting in the middle of a soccer field during a game. When the referee blows his whistle, one of the players r
NikAS [45]
Before its moving it should be 0 right 
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3 years ago
Read 2 more answers
Cart A of inertia m has attached to its front end a device that explodes when it hits anything, releasing a quantity of energy E
Leviafan [203]
We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
Let's start with energy conservation law:
\frac{mv^2}{2}+\frac{2mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+\frac{2mv_{B}^2}{2}
\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2
This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
mv+2mv=mv_A+2mv_B\\ 3mv=mv_A+2mv_B\\
Because both carts have the same mass we can cancel those out:
3v=v_A+2v_B
Now we have our system of equation that we have to solve:
\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\ 3v=v_A+2v_B
Part A
We need to solve our system for v_a. We will solve second equation for v_b and then plug that in the first equation.
3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=\frac{3v-v_A}{2}
Now we have to plug this in the first equation:
\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\
We will multiply the first equation with 2 and divide by m:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\
Now we plug in the second equation into first one:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+2\frac{(3v-v_A)^2}{4}\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+\frac{9v^2-6v\cdot v_A+v_{A}^2}{2} /\cdot 2\\ 6v^2+\frac{3E}{m}=2v_{A}^2+9v^2-6v\cdot v_A+v_{A}^2}\\ 3v_A^2-6v\cdot v_a+3(v^2-\frac{E}{m})=0/\cdot\frac{1}{3}\\ v_A^2-3v\cdot v_A+ (v^2-\frac{E}{m})=0
We end up with quadratic equation that we have to solve, I won't solve it by hand. 
Coefficients are:
a=1\\
b=-6v\\
c=v^2-\frac{E}{m}
Solutions are:
v_A=\frac{3v+\sqrt{5v^2+\frac{4E}{m}}}{2},\:v_A=\frac{3v-\sqrt{5v^2+\frac{4E}{m}}}{2}
Part B
We do the same thing here, but we must express v_a from momentum equation:
3v=v_A+2v_B\\
v_A=3v-2v_B
Now we plug this into our energy conservation equation:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_A={3v-v_B}\\
3v^2+\frac{3E}{2m}=(3v-v_B)^2+2v_B^2\\
3v^2+\frac{3E}{2m}=9v^2-6v\cdot v_B+v_B^2+2v_B^2\\
3v^2+\frac{3E}{2m}=3v_B^2-6v\cdot v_B+9v^2\\
3v_B^2-6v\cdot v_B+9v^2-3v^2-\frac{3E}{2m}=0\\
3v_B^2-6v\cdot v_B+(6v^2-\frac{3E}{2m})=0

Again we end up with quadratic equation. Coefficients are:
a=3\\
b=-6v\\
c=6v^2-\frac{3E}{2m}
Solutions are:
v_B=\frac{6v+\sqrt{-36v^2+\frac{18E}{m}}}{6},\:v_B=\frac{6v-\sqrt{-36v^2+\frac{18E}{m}}}{6}



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If a piece of wood dropped from a tall building has rezched a velocity of 68.6 m/s , how long has it been falling
notsponge [240]

It takes 7.0s for a piece of wood to be falling from a tall building in the velocity of 68.6 m/s

<u>Explanation:</u>

v = u + at

68.6 ms^-1 = 0 ms^-1 + (9.8 ms^-2 x t)

t = 68.6 ms^-1 / 9.8 ms^-2

t = 7.0s

It takes 7.0s for a piece of wood to be falling from a tall building in the velocity of 68.6 m/s

7 0
4 years ago
Three objects having the same mass begin at the same height, and all move down the same vertical distance H. One falls straight
Mnenie [13.5K]

Answer:

Work done in all the three cases will be the same.

Explanation:

1) The free falling body has only one force acting on it, the gravitational force. The work done on the body = mgH (Gravitational potential energy)

2) There are two forces acting on the body going down on a frictionless inclined plane - gravity and the normal force. The gravitational potential energy will be the same. The work done due to the normal force is zero, since the direction of the force is perpendicular to the displacement. Hence, total work done on the body = mgH

3) In the case of the body swinging on the end of a string, the change in gravitational potential enrgy will once again be the same since difference in height is H. The additional force on the body is the tension due to the string. But the work done due to this force is <em>zero, </em>since the displacement of the body is perpendicular to the tension. Therefore, the total work done on the body is once again mgH.

4 0
3 years ago
Starting from rest, a 75.0-kg snowboarder slides straight down a 115-m slope in 6.6 s. If the slope is a 35°
Likurg_2 [28]

The acceleration of the snowboarder is 5.6 m/s^2 down along the incline

Explanation:

To find the acceleration of the snowboard, we have to analyze the forces acting along the direction parallel to the incline.

There is only one force acting in this direction, and it is the component of the weight parallel to the incline, given by

mg sin \theta

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m = 75.0 kg is the mass of the man

g=9.8 m/s^2 is the acceleration of gravity

\theta=35^{\circ} is the angle of the incline

Substituting, we find that the net force along the incline is:

F=(75.0)(9.8)(sin 35^{\circ})=421.6 N

According to Newton's second law, the net force on the snowboard is equal to the product between his mass and his acceleration:

F=ma

where a is the acceleration. Therefore, solving for a, we find

a=\frac{F}{m}=\frac{421.6}{75}=5.6 m/s^2

And the direction of this acceleration is down along the incline.

Learn more about acceleration and Newton's second law:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
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