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3 years ago

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If a stone is dropped from a height of 400 feet, its height after t seconds is given by s = 400 − 16t2. Find its instantaneous v

elocity function.

1 answer:

soldier1979 [14.2K]3 years ago

5 0

Answer:

$v = -32 t$

Explanation:

given,

s = 400- 16 t²

we know,

Velocity of an object is defined as the change in displacement per unit change in time.

velocity an also be return as

$v = \dfrac{ds}{dt}$

$v = \dfrac{d}{dt}(400-16t^2)$

$v= 0 -2\times 16 t$

$v = -32 t$

Hence, instantaneous velocity function given by $v = -32 t$

To calculate instantaneous velocity, you need to insert value of time.

ex, instantaneous velocity at t = 4 s

v = -32 x 4 = -128 m/s.

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Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with

Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

For the first stone

$y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}$

$0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}$

$0=y_{i1}+8t_{1}-4.905*t_{1}^{2}$ (1)

For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

<u>Part B)</u>

We can use the equation of the horizontal position here.

<u>First stone</u>

$x_{f1}=x_{i1}+vcos(30)t_{1}$

$x_{f1}=0+13.86*t_{1}$

$x_{f1}=13.86*t_{1}$

<u>Second stone</u>

$x_{2}=x_{i2}+vcos(-30)t_{2}$

$x_{1}=0+13.86*t_{1}$

$x_{1}=13.86*t_{2}$

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

3 0

3 years ago

On a trolley ride around an amusement park, a child travelled from one signpost to a second signpost at a constant speed of 125

Tomtit [17]

Answer:

Explanation:

Speed given = 125 m /min

125 /60 m /s

In 450 second it will travel

= 450 x 125 / 60

=937.5 m.

As the distance is covered in less than 450 seconds , The distance must be less than 937.5 m

In 400 seconds , it will travel

= 400 x 125 / 60

833.33 m

Since the distance is covered in more than 400 seconds , the distance must be more than ie 833.33 .

Hence the distance covered is more than .833 m but less than 937.5

In either case these distance are more than .8 km .

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3 years ago

How much potential energy does a 40n block medicine ball gain when it is lifted 5.m

MAVERICK [17]

By definition, the gain in PE (potential energy) is
ΔPE = m*g*h

Given:
mg = 40 N (Note that m*g = weight)
h = 5 m

ΔPE = (40 N)*(5 m) =200 J

Answer: 200 J

5 0

3 years ago

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A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se

zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, $\lambda = 1\ nC = 1\times 10^{- 9}\ C$

Charge density of rod 2, $\lambda = - 1\ nC = - 1\times 10^{- 9}\ C$

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

$\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}$

Also,

$\vec{E} = \frac{2K\lambda }{R}$ (1)

where

K = electrostatic constant = $\frac{1}{4\pi \epsilon_{o} R}$

R = Distance

$\lambda$ = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

$\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C$

$\vec{E} = 1800\ N/C$ (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

$\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C$

$\vec{E'} = 1800\ N/C$ (towards)

Now, the total field at the origin is the sum of both the fields:

$\vec{E_{net}} = 1800 + 1800 = 3600\ N/C$

7 0

3 years ago

What is displacemnet

allsm [11]

Answer:

to move an object from its place or position

Explanation:

if you move an object from where it first was

7 0

3 years ago

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