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zloy xaker [14]
3 years ago
5

When a 58g tennis ball is served, it accelerates from rest to a constant speed of 36 m/s. The impact with the racket gives the b

all a constant acceleration over a distance of 35 cm. What is the magnitude of the net force acting on the ball?
Physics
1 answer:
inysia [295]3 years ago
5 0
We first calculate the acceleration on the ball using:
2as = v² - u²; u = 0 because ball is initially at rest
a = (36)²/(2 x  0.35)
a = 1850 m/s²
F = ma
F = 0.058 x 1850
= 107.3 Newtons
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igomit [66]

Answer: Linear speed is 1,670 Kph.

Explanation:

If we assume that the earth is a perfect sphere, and that is spinning itself once every roughly 24 hr, we can get the angular velocity of the Earth, in magnitude, as follows:

ω = 2π / 24 Hr

Now, by definition, an angle is the relationship between the arc s, and the radius r, so we can replace these values in the angular velocity expression, as follows:

ω = (Δs / r) . 1/Δt ⇒ ω = (Δs/Δt). 1/r

But, by definition, Δs/At, is just the linear velocity, v, so we can conclude the following;

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4 0
3 years ago
A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a
Rasek [7]

Answer:

W=173.48J

Explanation:

information we know:

Total force: F=45N

Weight: w=100N

distance: 4m

vertical component of the force: F_{y}=12N

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: F_{x}=Fcos\theta

vertical component: F_{y}=Fsen\theta

but from the given information we know that F_{y}=12N

so, equation these two F_{y}=Fsen\theta and F_{y}=12N

Fsen\theta =12N

and we know the force F=45N, thus:

45sen\theta=12

now we clear for \theta

sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

F_{x}=Fcos\theta

F_{x}=45cos(15.466)\\F_{x}=43.37N

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J

the work done is W=173.48J

7 0
3 years ago
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konstantin123 [22]

Answer:

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4 0
3 years ago
What's something that I've done in my life that is related to physics
Allisa [31]

That's a very difficult question to answer, because you give us
no information regarding what you have done in your life. 

We can only assume that you have most likely breathed on occasion,
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ladder or out of bed, felt the warmth of the sun on your cheek, seen
a rainbow after a rainshower, heard the sound of thunder during a
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6 0
3 years ago
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One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

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Where,

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T_H = Hot spot temperature

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T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

\eta = 1-\frac{T_L}{T_H}

\eta = 1-\frac{293}{713}

\eta = 0.589

Therefore the maximum possible efficiency the car can have is 58.9%

4 0
3 years ago
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