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Andreas93 [3]
3 years ago
12

A ball rolls of buildings that is 100m high calculate the time that it takes for ball to hit the ground​

Physics
1 answer:
LUCKY_DIMON [66]3 years ago
5 0

Answer:

2as=v2-u2

2000=v2

V=44

V=u+at

44/10=t

T=4.4seconds

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The record for the highest speed achieved in alaboratory for a
Ira Lisetskai [31]

Answer:

0.000234 seconds

Explanation:

Since the row is 0.15m, its radius of rotation must be 0.15 / 2 = 0.075 m

We can start by calculating the angular speed of the rod:

\omega = \frac{v}{r} = \frac{2010}{0.075} = 26800 rad/s

Since one revolution equals to 2π rad. The speed in revolution per second must be

26800 / 2π = 4265 revolution/s

The number of seconds per revolution, or period, is the inverse:

1/4265 = 0.000234 seconds

3 0
3 years ago
5. The net external force on a rock of mass 4.2 kg is 8.0 N forward. Find the acceleration of the rock.
andreev551 [17]

Answer:

1.904

Explanation:

F= ma

8 = 4.2 a

a = 8/4.2

a = 1.904

6 0
2 years ago
Now consider a wave which is paired with seven other waves into seven pairs. The two waves in each pairing are identical, except
kupik [55]

The pair BCEG will interfere constructively, while the pair ADF will interfere destructively.

Constructive and destructive interference:

For interference, the waves must be coherent.

Two coherent waves interfere constructively when the path difference is equal to an integral multiple of the wavelength.

That is the path difference must be mλ

where m = 0,1,2,3.... is an integer and λ is the wavelength

So pair BCEG interfere constructively

Two coherent waves interfere destructively when the path difference is equal to a half-integral multiple of the wavelength.

That is the path difference must be (m+1/2)λ

where m = 0,1,2,3....   is an integer and λ is the wavelength

Therefore, the interference is in the pair ADF which is interfere destructively.

Learn more about interference: brainly.com/question/1194044

#SPJ4

[NOTE: THE COMPLETE QUESTION IS:

Now consider a wave which is paired with seven other waves into seven pairs. The two waves in each pairing are identical, except that one of them is shifted relative to the other in the pair by the distance shown: A. -(1/2) ?B. 2?C. -5?D. (3/2)?E. 0F. (17/2)?G. (6/2)?Identify which of the seven pairs will interfere constructively and which will interfere destructively. Each letter represents a pair of waves. Enter the letters of the pairs that correspond to constructive interference in alphabetical order and the letters of the pairs that correspond to pairs that interfere destructively in alphabetical order separated by a comma. For example if pairs A, B and D interfere constructively and pairs C and F interfere destructively enter ABD,CF.]

4 0
1 year ago
Read 2 more answers
If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in
Sindrei [870]

Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm

5 0
3 years ago
José and Laurel measured the length of a stick's shadow during the day. Without knowing the length of the stick, which of their
skelet666 [1.2K]

Answer:

Option B.

Explanation:

Assuming the stick is in vertical position, its shadow depends on two factors: its length and the angle between the sun rays and the stick. When the angle is bigger, the lenght of the shadow increases, and vice versa. So, when the sun rays are parallel to the stick, the shadow may be small. Since they are nearly perpendicular to the Earth's surface at 12 o'clock, the shadow of the stick at that time should be minimal. It means that the measured shadow of 75 cm at 12:30 p.m. is almost impossible (Option B).

5 0
3 years ago
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