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12345 [234]
3 years ago
7

You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r

esistance)?
Physics
1 answer:
Brut [27]3 years ago
8 0

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

d=ut-\dfrac{1}{2}gt^2

Putting all values

d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

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Interactive Solution 11.13 presents a model for solving this problem. A solid concrete block weighs 100 N and is resting on the
mash [69]

Answer:

The value is }  N  =  66 \  blocks

Explanation:

From the question we are told that

The weight of the block is W_b  = 100 \  N

The dimension of the block is d =  0.400 m  \ X  \ 0.250 \  m  \  X  \ 0.130 \ m

Generally two atmosphere is equivalent to

P_{2atm} =  2 *  1.013 *10^{5} =  202600 \  N/m^2

Generally 1 atm = 1.013 *10^{5} N/m^2

The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks

So

A =  0.250 *  0.130

=> A =  0.0325 \  m^2

Generally the force due to this blocks is mathematically represented as

F =  N  *  W_b

Here N is the number of blocks

So

}  202600 =  \frac{N  *  100 }{ 0.0325}

=>   }  N  =  66 \  blocks

3 0
3 years ago
A body starts from rest and travels ‘s’ m in 2nd second, than acceleration is.
Lana71 [14]

Answer:

3m

Explanation:

3 0
2 years ago
What can be defined as the rate at which velocity changes
taurus [48]

Answer:

acceleration

Explanation:

The rate at which velocity changes is the definition of the physical quantity called acceleration, and it is given by the formula: a=\frac{v_f-v_i}{\Delta t}

where \Delta t is the time that took to change from the initial velocity v_i to the final velocity v_f

4 0
3 years ago
Read 2 more answers
Compare and contrast series and parallel circuits?
dalvyx [7]

In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.

Hope It Helps!

6 0
2 years ago
A mass weighting 48 lbs stretches a spring 6 inches. The mass is in a medium that exerts a viscous resistance of 27 lbs when the
Mademuasel [1]

Answer:

a)

u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

b)

m = 48lb

c)

b = 144.76lb

Explanation:

The general equation of a damping oscillate motion is given by:

u(t)=u_oe^{-\frac{b}{2m}t}cos(\omega t-\alpha)    (1)

uo: initial position

m: mass of the block

b: damping coefficient

w: angular frequency

α: initial phase

a. With the information given in the statement you replace the values of the parameters in (1). But first, you calculate the constant b by using the information about the viscous resistance force:

|F_{vis}|=bv\\\\b=\frac{|F_{vis}|}{v}\\\\|F_{vis}|=27lbs=27*32.17ft.lb/s^2=868.59ft.lb/s^2\\\\b=\frac{868.59}{6}lb/s=144.76lb/s

Then, you obtain by replacing in (1):

6in = 0.499 ft

u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

b.

mass, m = 48lb

c.

b = 144.76 lb/s

8 0
3 years ago
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