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I am Lyosha [343]
2 years ago
9

Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. What is the approxi

mate horizontal component of the initial velocity?
Physics
2 answers:
AleksandrR [38]2 years ago
7 0
The horizontal component is calculated as:
Vhorizontal = V · cos(angle)

In your case Vhoriontal = 16 · cos(40) = 12.3 m/s

Answer: 12.3 m/s
marysya [2.9K]2 years ago
4 0

Answer:

Answer:D 12.3 m/s

Explanation:

I just took the quiz

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An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t
gtnhenbr [62]

1. tan \theta = \frac{v^2}{Rg}

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

N cos \theta = mg (1)

N sin \theta = m \frac{v^2}{R} (2)

where in the second equation, the term m\frac{v^2}{R} represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the  following expression:

tan \theta = \frac{v^2}{Rg}

2. F=\frac{m}{R}(w^2-v^2)

In this second situation, the cars moves around the track at a speed

w>v

This means that the centripetal force term

m\frac{v^2}{R}

is now larger than before, and therefore, the horizontal component of the normal reaction, N sin \theta, is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

N sin \theta + F = m\frac{w^2}{R}

And re-arranging for F,

F=m\frac{w^2}{R}-N sin \theta (3)

But from eq.(2) in the previous part we know that

N sin \theta = m \frac{v^2}{R}

So, susbtituting into eq.(3),

F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)

4 0
2 years ago
1 Ten (10) ml aqueous solutions of drug A (10% w/v) and drug B (25% w/v) are stored in two identical test tubes under identical
Reil [10]

Answer:

YOUR answer is given below:

Explanation:

3 0
1 year ago
A 0.45 Caliber bullet (m = 0.162 kg) leaving the muzzle of a gun at 860
Charra [1.4K]

Answer:139.32kgm/s

Explanation:

Mass(m)=0.162kg

Velocity(v)=860m/s

Momentum=mass x velocity

Momentum=0.162 x 860

Momentum=139.32kgm/s

3 0
1 year ago
A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.76 m and whose unstretched length is 0.41 m. Next the
Blizzard [7]
<h3><u>Answer;</u></h3>

= 1.256 m

<h3><u>Explanation;</u></h3>

We can start by finding the spring constant  

F = k*y  

Therefore;  k = F/y = m*g/y

                               = 0.40kg*9.8m/s^2/(0.76 - 0.41)

                               = 11.2 N/m  

Energy is conserved  

Let A be the maximum displacement  

Therefore;  1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2  

Thus;  A = sqrt((1.20 - 0.55)^2 + m/k*v^2)

               = sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)

                = 0.846 m  

Thus; the length will be 0.41 + 0.846  = 1.256 m

6 0
2 years ago
the initial kinetic energy of an object moving on a horizontal surface is K. Friction between the object and the surface causes
Ugo [173]

Answer:

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1 year ago
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