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Eduardwww [97]
3 years ago
12

A ball is thrown straight up with enough speed so that it is in the air for several seconds. Assume the positive direction is up

wards. Part APart complete What is the velocity of the ball when it reaches its highest point? Express your answer to two significant figures and include the appropriate units. v = 0 ms Previous Answers Correct Part B What is its velocity 0.70 s before it reaches its highest point?
Physics
1 answer:
Andrew [12]3 years ago
6 0

Answer:

a) v= 0 m/s b) v= 6.86 m/s

Explanation:

a) When the ball reaches to its highest point, under the influence of gravity, before starting to fall down, it momentarily comes to an stop (this is needed prior to change direction in any movement), so, applying the definition of acceleration, and replacing the acceleration a by g, we have:

vf = v₀ - g*t (1)

The minus sign means that the acceleration due to gravity is always downward, so if we assume that the positive direction is upwards it must be negative.

At the highest point, vf= 0.

b) Prior to solve this point, we need to know which is the time when the ball reaches to its highest point.

As we know vf=0, we can solve (1) for t, as follows:

th = v₀/g

Now, for a time that is 0.7 s before this time, applying the acceleration definition and solving for v again, we have:

v = v₀ -(g *(th-0.7 s)), but th= v₀/g, so we get:

v= v₀ -g((v₀/g)-0.7 s) = v₀ - v₀ + g*0.7 s

⇒ v=g*0.7 s = 9.8 m/s²*0.7 s

⇒ v = 6.86 m/s

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When an object is turning a corner what direction is the acceleration?
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If the corner is rounded and is perfectly circular, then the acceleration is centripetal and is always directed toward the center.
7 0
3 years ago
A spool of thread has an average radius of 1.00 cm. If the spool contains 62.8 m of thread, how many turns of thread are on the
Simora [160]

To solve this problem we will start from the given concept in which the number of turns is equivalent to the length of the thread per circumference of spool. That is:

N = \frac{l}{\phi}

Where,

l = length of the thread

\phi= circumference of spool

For \phi we have that,

\phi = 2\pi r \rightarrow 2\pi (0.01)

For l  we have that

l = 62.8m

Finally the number of Turns would be,

N = \frac{l}{\phi}

N = \frac{62.8}{2\pi (0.01)}

N = 1000turns

Therefore the number of turns of thread on the spool are 1000turns.

7 0
3 years ago
Your car gets a flat! You go from 90 kilometers per hour to a stop in 6 seconds. What is your rate of deceleration?
Gelneren [198K]

Answer:

The rate of deceleration is 15 km/h^2.

Explanation:

Since the car comes in 90km/h, it will need deceleration for 15 km/h for 6 seconds to finally stop the car. 15*6=90

8 0
2 years ago
A certain freely falling object, released from rest, requires 1.50 s to travel the last 30.0 m before it hits the ground. From w
BaLLatris [955]

Answer:

The object fell from about 38.14 meters

Explanation:

We can use the formula for displacement under accelerated motion due to gravity to find the velocity of the object had 30 m before hitting the ground :

y_f-y_i=v_i * \,t-\frac{9.8}{2} \,t^2\\0-30=v_i \,(1.5)-4.9\,(1.5)^2\\-18.975=v_i\,(1.5)\\v_i=-12.65\,\,\frac{m}{s}

Now, knowing this velocity, we can find the time it took to fall from the initial position to 30 m before hitting the ground:

v_f=v_i-g\,*\,t\\-12.65=0-g \,*\,t\\t=\frac{12.65}{g} \\t\approx1.29\,\,s

And now we can find what is the total distance covered in 1.5 s plus 1.29 seconds for this free falling object:

y_f-y_i=v_i * \,t-\frac{9.8}{2} \,t^2\\0-y_i=0-\frac{9.8}{2} \,(2.79)^2\\-y_i=-38.14\\y_i=38.14\,\,m

7 0
2 years ago
If a 200 turn, 10-3 m2 cross-sectional area coil is immersed in a magnetic field such that the plane of the coil is perpendicula
Stels [109]

Answer:

<h3>1.2 volt induced in coil.</h3>

Explanation:

Given:

Number of turns N =  200

Cross sectional area A = 10^{-3} m^{2}

Rate of increasing magnetic field \frac{dB}{dt} = 6 \frac{T}{s}

From the faraday's law,

Induced emf is given by,

   \epsilon = -N\frac{d \phi}{dt }

Where \phi = magnetic flux

  \phi = AdB \cos(0)          ( because angle between normal coil and field is zero)

Where A = area of coil

Put the value of \phi in above equation,

Here we neglect minus sign

   \epsilon = NA\frac{dB}{dt}

   \epsilon = 200 \times 10^{-3} \times 6

   \epsilon = 1.2 V

Therefore, 1.2 volt induced in coil

8 0
3 years ago
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