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2 years ago

10

Why can’t you hit a feather in mid air with a force of 200 N (This is all the info I was given)

1 answer:

Aloiza [94]2 years ago

5 0

You cannot for real ehhhh

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If we know the total energy in a system is 30 J, and we know the PE is 20 J. What is the KE?

svetoff [14.1K]

Can you give more description ??

3 0

3 years ago

If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i

kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from $15 cm$ to $21 cm$ will be , $E_{1}=27J$

The energy used for stretch the spring from $21 cm$ to $27 cm$ will be , $E_{2}=45J$

using the energy of spring formula ,we find that

$27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))$

$45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))$

Dividing both the equation will get,

$\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm$

Therefore, the natural length of the spring is, 9 cm.

4 0

2 years ago

Car A is traveling west at 40 mi/h and car B is traveling north at 40 mi/h. Both are headed for the intersection of the two road

Gwar [14]

Explanation:

It is given that,

$\frac{dx}{dt}$ = -40 mi/h, $\frac{dx}{dt}$ = -40 mi/h

The negative sign indicates that x and y are decreasing.

We have to find $\frac{dz}{dt}$ . Equation for the given variables according to the Pythagoras theorem is as follows.

$z^{2} = x^{2} + y^{2}$

Now, we will differentiate each side w.r.t 't' as follows.

$2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$

or, $\frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})$

So, when x = 4 mi, and y = 3 mi then z = 5 mi.

As, $\frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})$

= $\frac{1}{5}(4 \times (-40) + 3 \times (-40))$

= $\frac{-140 - 120}{5}$

= 52

Thus, we can conclude that the cars are approaching at a rate of 52 mi/h.

7 0

3 years ago

A cannon ball is shot straight upward with a velocity of 72.50 m/s. How high is the cannon ball above the ground 3.30 seconds af

disa [49]

Answer:

Explanation:

Given

Cannon is fired with a velocity of $u=72.50\ m/s$

Using Equation of motion

$y=ut+\frac{1}{2}at^2$

where

$y=displacement$

$u=initial\ velocity$

$a=acceleration$

$t=time$

after time $t=3.3 s$

$y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2$

$y=239.25-53.36$

$y=185.89\ m$

So after 3.3 s cannon ball is at a height of 185.89 m

6 0

3 years ago

If it takes 50m to stop a car initially moving at 25mps, what distance is required to stop a car moving 50mps?

Anastaziya [24]

100m... Hope this helps :)

5 0

3 years ago

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