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Romashka-Z-Leto [24]
2 years ago
5

a horse gallops a distance of 60 meters in 15 seconds. then, he stops to eat some grass for 20 seconds. next, he trots for 25 se

conds over 60 meters. (the horse trots in the same direction he galloped.) finally, the horse races home (back to its starting position), traveling 120 meters in 20 seconds. graph the horse’s movements on the following distance-time chart.

Physics
1 answer:
Neporo4naja [7]2 years ago
3 0

Answer: check the attached graph for the answer

Explanation: Please find the attached file for the solution

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If the intensity level by 15 identical engines in a garage is 100 dB, what is the intensity level generated by each one of these
insens350 [35]

To develop this problem it is necessary to apply the concepts related to Sound Intensity.

By definition the intensity is given by the equation

\beta = 10Log(\frac{I}{I_0})

Where,

I = Intensity of Sound

I_0= Intensity of Reference

At this case we have that 15 engines produces 15 times the reference intensity, that is

I= 15I_0

And the total mutual intensity is 100 dB, so we should

\beta = 100-10*log(\frac{15I_0}{I_0})

\beta = 100-10*log(15)

\beta = 100-11.76

\beta = 88.23dB

Therefore each one of these engines produce D. 88dB.

5 0
2 years ago
If a pizza delivery guy declares himself the leader over the other pizza delivery people, he probably won't have many people obe
NemiM [27]

Answer:

the last one i think...

4 0
2 years ago
Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley
kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = \frac{g}{7}

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

Explanation:

Given data:

mass suspended = 4 meters

mass suspended at other end = 3 meters

first we have to express the kinetic and potential energy equations

The general kinetic energy of the system can be written as

T = \frac{4m}{2} x^2  + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2

T = 4mx^2 + 2my^2 -2mxy  

also the general potential energy can be expressed as

U = -4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant

The Lagrangian of the problem can now be setup as

L =4mx^2 +2my^2 -2mxy +2mgy + constant

next we will take the Euler-Lagrange equation for the generalized equations :

Euler-Lagrange  equation = 4x-y =0\\-2y+x +g = 0

solving the equations simultaneously

x ( acceleration of mass 4m ) = \frac{g}{7}

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

8 0
2 years ago
You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A
garri49 [273]

Answer:

The speed with which the baseball leaves the hand = 20.58 m/s

Explanation:

The time take to reach highest height during a projectile's flight is given by

t = (u sin θ)/g

u = initial velocity of the baseball = ?

θ = angle of throw above the horizontal

g = acceleration due to gravity = 9.8 m/s²

1.05 = (u sin 30)/9.8

u = (1.05 × 9.8)/0.5

u = 20.58 m/s

7 0
2 years ago
What force would be needed to accelerate a 0.040-kg golf ball at 20.0 m/s?
Naily [24]

Answer:

any amount of force will do it as time is not mentioned here

5 0
2 years ago
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