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3 years ago

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The runner reaches the first camera, at the 500 m mark, at 9:03:05 a.m. and the second camera, at the 600 m mark, at 9:03:25 a.m

. What is her average speed over that time interval

1 answer:

SVEN [57.7K]3 years ago

7 0

Answer:

Explanation:

Distance travelled = 600 - 500 = 100 m

Time taken = 9:03:25 a.m.- 9:03 :05 a.m.

= 20 s

average speed = distance / time taken

= 100 / 20 m /s

= 5 m /s

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You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen

Paladinen [302]

Answer:

(A)

$\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s \right )D+t_t^2v_s^2=0$

<em>(B) D=54.71 m</em>

Explanation:

<u>Free Fall</u>

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

$\displaystyle D=\frac{gt^2}{2}$

Solving for t

$\displaystyle t=\sqrt{\frac{2D}{g}}$

If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

$\displaystyle t=\frac{D}{v_s}$

(A)

The total measured time is the sum of both times and it's given as $t_t=3.5\ seconds$

$\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}$

From this equation we'll manage to compute D

First, we isolate the square root

$\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}$

Let's square both sides

$\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}$

Multiplying by $v_s^2$

$\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2$

Rearranging and factoring

$\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}$

Now, let's put in numbers:

$g=9.8\ m/s^2,\ v_s=345\ m/s,t_t=3.5\ sec$

$\displaystyle D^2-\left (\frac{2(345)^2}{9.8}+2(3.5)(345)\right )D+(12.25)345^2=0$

Computing all the coefficients:

$\displaystyle D^2-26,705.82D+1,458,056.25=0$

Solving for D, we have two possible solutions:

$D=54.71,\ D=26,651.11$

The second solution is called "extraneous", since it comes from squaring an equation, which can introduce non-valid (or external) solutions. It's impossible, given the conditions of the problem, that the well could be 26.5 km deep. So we'll keep the only solution as.

<em>D=54.71 m</em>

Let's prove our calculations by computing both times:

$\displaystyle t_1=\sqrt{\frac{2(54.71)}{9.8}}=3.34\ sec$

$\displaystyle t_2=\frac{54.71}{345}=0.16\ sec$

We can see their sum is 3.5 seconds, 3.34 of which were taken to reach the bottom of the well, and 0.16 sec took the sound to reach the top.

3 0

3 years ago

What gravitational force does the moon produce

saul85 [17]

Answer:

$1.94\cdot 10^{20} N$

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

The gravitational force is always attractive.

In this problem, we have:

$m_1 = 5.98\cdot 10^{24}kg$ is the mass of the Earth

$m_2 = 7.34\cdot 10^{22} kg$ is the mass of the Moon

$r=3.88\cdot 10^8 m$ is the separation between the Earth and the Moon

Therefore, the gravitational force between them is

$F=(6.67\cdot 10^{-11})\frac{(5.98\cdot 10^{24})(7.34\cdot 10^{22})}{(3.88\cdot 10^8)^2}=1.94\cdot 10^{20} N$

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3 years ago

Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergo

oksian1 [2.3K]

Answer:

e). $a = 0.066 m/s^2$

Explanation:

As we know that wheel is turned by 90 degree angle

so the angular speed of the wheel is given as

$\omega_f^2 - \omega_i^2 = 2\alpha \theta$

now we have

$\omega_f^2 - 0 = 2(0.01)(\frac{\pi}{2})$

$\omega = 0.177 rad/s$

now the centripetal acceleration of the point P is given as

$a_c = \omega^2 R$

$a_c = (0.177)^2(2)$

$a_c = 0.063 m/s^2$

tangential acceleration is given as

$a_t = R\alpha$

$a_t = 2(0.01)$

$a_t = 0.02 m/s^2$

now net acceleration is given as

$a = \sqrt{a_t^2 + a_c^2}$

$a = \sqrt{0.02^2 + 0.063^2}$

$a = 0.066 m/s^2$

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3 years ago

If an airplane is flying at 300 km/h to the east and is facing a headwind of 18.0 km/h, what is the airplane's final velocity?

givi [52]

If an airplane is flying at 300 km/h to the east and is facing a headwind of 18.0 km/h, the final velocity can be calculated using simple vector addition. In this case, the planes velocity is positive (+330 km/h) and head wind has a negative component (-18.0 km/h). Vector addition yields +330 km / h + (-18.0 km /h) = 312 km / h.

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joja [24]

Answer:

Explanation:

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