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2 years ago

A ball dropped from a window strikes the ground 2.71 seconds later. How high is the window above the ground?

Physics

1 answer:

koban [17]2 years ago

4 0

$s=36.52m$

Given ,

Time =2.71 second

We apply the formula

$s=ut+12at^2$

$u=0,t=2.73 \\a=g \\So, the \\height \\ s=36.52m$

When a body is lying on a horizontal surface, the object is subjected to a downward force due to the action of gravity. These gravitational forces always aim to attract the item, confirming the force's nature as an attractive force. Consider an object of mass "M" resting on a horizontal surface; the force exerted to the object by gravity is equal to the weight of the object "W." The acceleration owing to gravity, indicated as "g," is the acceleration by which an object is drawn towards the ground. As a result, Newton's second law states that when the values in the above connection are substituted.

To learn more about Motion due to gravity from the given link:

brainly.com/question/14047086

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A bus decreases its speed

eduard

Answer:

-6 km/h per minute

Explanation:

7 0

3 years ago

A submarine is 2.84 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the

xxMikexx [17]

Answer:

468 m

Explanation:

So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.

The length CE is length AE - length AC = 284 - 234 = 50 m

We can calculate the angle ECD:

$tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2$

$\hat{ECD} = tan^{-1} 2 = 63.43^o$

This is also the angle ACB, so we can find the length AB:

$tan(\hat{ACB}) = \frac{AB}{AC} = \frac{AB}{234}$

$2 = \frac{AB}{234}$

$AB = 2*234 = 468 m$

So the height of the building is 468m

5 0

4 years ago

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

3 years ago

Which type of interference occurs when two waves exactly cancel out? NEED AN ANSWER ASAP

Nostrana [21]

If the two waves combine to produce ANY wave that smaller
than either of the originals, that's destructive interference.

4 0

3 years ago

Chinook salmon are able to move upstream faster by jumping out of the water periodically; this behavior is called porpoising. Su

deff fn [24]

Answer:he formula for average speed is (total distance/total time)

the y-component does not matter in this problem. so do 6.26(cos45)=4.43m/s to find the x-component velocity which is constant throughout the duration of the flight. the total distance is 2L because he travels distance L twice.

the total time is ((time in water)+(time out of water)) since you dont have time you must eliminate it. to do this you need (distance)/(time)=velocity

solve for time and you get T=D/V

time in water is L/3.52 and time out of water is L/4.43

add them together and you get (4.43L+3.52L)/(15.59) = 7.95L/15.59

that value is your total time

divide you total distance (2L) by total time (7.95L/15.59) and the Ls cancel out and you get

(31.18)/(7.95) = 3.92 m/s = Average Speed

Explanation:

7 0

4 years ago