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3 years ago

Andy took a bus and then walked from his home to downtown.

Physics

1 answer:

Sunny_sXe [5.5K]3 years ago

5 0

Answer:

12.9

Explanation:

1.6x15=24

0.4x4.5=1.8

24+1.8=25.8

1.6+0.4=2

25.8/2=12.9

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Based on Schaffer's arguments in "The Value of a Sherpa Life," choose the statement

Tju [1.3M]

Answer:

Sherpas do work that is much more meaningful than the work other climbers do.

Explanation:

3 0

3 years ago

The property of water that contributes to its ability to stick to certain surfaces is called

Gekata [30.6K]

Answer: Adhesion

Explanation: Adhesion is the tendency of dissimilar particles or surfaces to cling to one another (cohesion refers to the tendency of similar or identical particles/surfaces to cling to one another). The forces that cause adhesion and cohesion can be divided into several types. This allows Particles in things like water to stick to surfaces

3 0

3 years ago

A law enforcement officer in an intergalactic "police car" turns on a red flashing light and sees it generate a flash every 1.2

butalik [34]

Answer:

The velocity of the police car relative to earth is $v_{rel} = 2.51\times 10^{8} m/s$

Given:

time for flash generation of the inter galactic police car, t = 1.2 s

time between flashes as measured from earth, t' = 2.2 s

Solution:

Utilising Einstein's equation for time dilation to calculate the velocity of the police car, the equation is given by:

$t' = \frac{t}{\sqrt {1 - \frac{v^{2}}{c^{2}}}}$ (1)

where, c = speed of light in vacuum = $c = 3\times 10^{8}$

re arranging eqn (1) for velocity, v:

$v_{rel} = c\times \sqrt {1 - (\frac{t}{t'})^{2}}$ (2)

Now, from eqn (2)

$v_{rel} = 3\times 10^{8}( \sqrt {1 - (\frac{1.2}{2.2})^{2}})$

$v_{rel} = 3\times 10^{8}\times 0.838$

$v_{rel} = 2.51\times 10^{8} m/s$

3 0

3 years ago

What is the minimum speed must a salmon jumping at an angle of 35.2 leave the water in m/s?

yuradex [85]

Height of the waterfall is 0.449 m

its horizontal distance will be 2.1 m

now let say his speed is v with which he jumped out so here the two components of his velocity will be

$v_x = vcos35.2 = 0.817 v$

$v_y = vsin35.2 = 0.576 v$

here the acceleration due to gravity is 9.81 m/s^2 downwards

now we can find the time to reach the other end by y direction displacement equation

$\Delta y = v_y * t + \frac{1}{2} at^2$

$-0.449 = 0.576 * v *t - \frac{1}{2}*9.81 * t^2$

also from x direction we can say

$\Delta x = v_x * t$

$2.1 = 0.817 v* t$

now we have

$v* t = 2.57$

we will plug in this value into first equation

$- 0.449 = 0.576 * 2.57 - 4.905 * t^2$

$1.93 = 4.905 * t^2$

$t = 0.63 s$

now as we know that

$v* t = 2.57$

t = 0.63 s

$v = \frac{2.57}{0.63}$

$v = 4.1 m/s$

so his minimum speed of jump is 4.1 m/s

8 0

3 years ago

If a box of supplies is dropped from the cargo hold of an airplane travelling at an altitude of 4,410 meters how long will it ta

Paraphin [41]

Answer:

It will take 30 seconds to reach the ground, and it will be travelling at 294 m/s when it does so. This means that its average velocity was 147 m/s.

Explanation:

$d=v_ot+\dfrac{1}{2}at^2$

Since the initial velocity of a dropped object is 0, we can make this the equation:

$d=\dfrac{1}{2}at^2 \\\\4410=\dfac{1}{2}(9.8)t^2 \\\\t^2=900$

$t=30\text{ seconds}$

The final velocity can be calculated with the formula:

$v_f=v_o+at$

Once again, since there is no initial velocity:

$v_f=at \\\\v_f=(9.8)(30)=294m/s$

Since the initial velocity is 0, the average vertical velocity is 294/2=147 m/s.

Hope this helps!

5 0

3 years ago