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gregori [183]
3 years ago
11

Calculate the force of friction that keeps an 80-kg person sitting on the edge of a horizontal rotating platform when the person

sits 2 m from the center of the platform and has a tangential speed of 3 m/s.
Physics
1 answer:
Lilit [14]3 years ago
8 0

Answer:

360N

Explanation:

The frictional force = centripetal force = mass * centripetal acceleration = ma

Since centripetal acceleration = V^2/ r

Frictional force = mv^2 / r

Substitute the values of m = 80kg, radius = 2m and tangential speed = 3m/s

Frictional force = 80 * (3)^2/ 2 = 360N

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Boston Red Sox pitcher Roger Clemens could routinely throw a fastball at a horizontal
azamat

Answer: 0.145 seconds

Explanation:

Given that Roger Clemens could routinely throw a fastball at a horizontal speed of 119.7 m/s. How long did the ball take to reach home plate 17.3 m away

Since the speed is horizontal

Using the formula for speed which is

Speed = distance/time

Where

Speed = 119.7 m/s

Distance covered = 17.3 m

Time is what we are looking for

Substitute all the parameters into the formula

119.7 = 17.3/ time

Make time the subject of formula

Time = 17.3 / 119.7

Time = 0.145 seconds.

Therefore, it will take 0.145 seconds to reach the home plates

6 0
2 years ago
The chemical symbol for sulfuric acid is H2SO4. How many atoms are contained in each molecule of sulfuric acid?
cricket20 [7]
You know from looking at the molecular formula<span> that one </span>molecule<span> of </span>H2SO4<span> contains 2 </span>atoms<span> of hydrogen, 1 atom of sulfur and 4 </span>atoms<span> of oxygen.</span>
5 0
3 years ago
What are the 3 formulas you can use for vertical motion for a projectile?
klio [65]
I hope this can help you ask me if you need help again

4 0
3 years ago
A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin
Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

3 0
3 years ago
A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Serggg [28]

Answer:

The spring force constant is  k=243\ \frac{N}{m} .

Explanation:

We are told the mass of the ball is m=0.0555\ kg, the height above the spring where the ball is dropped is h=0.536\ m,  the length the ball compresses the spring is d=0.04897\ m and the acceleration of gravity is 9.8\ \frac{m}{s^{2}} .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy E_{mi} is equal to the final mechanical energy E_{mf} :

                                                    E_{mf}=E_{mi}

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

                                                   \frac{1}{2}kd^{2}=mgh

If we manipulate the equation we have that:

                                                    k=\frac{2mgh}{d^{2} }

                                         k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}

                                              k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}

                                                     k=243\ \frac{N}{m}

                                                   

                             

5 0
3 years ago
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