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Marina86 [1]
4 years ago
14

While standing on a balcony a child drops a penny. The penny lands on the ground floor 1.5 s later. How fast was the penny trave

ling vertically when it struck the ground floor?
*The height of the balcony is 11.025 meters

​

Physics
1 answer:
sveticcg [70]4 years ago
8 0

Answer:

14.7 m/s.

Explanation:

From the question given above, the following data were obtained:

Time (t) = 1.5 s

Acceleration due to gravity (g) = 9.8 m/s².

Height = 11.025 m

Final velocity (v) = 0 m/s

Initial velocity (u) =?

We, can obtain the initial velocity of the penny as follow:

H = ½(v + u) t

11.025 = ½ (0 + u) × 1.5

11.025 = ½ × u × 1.5

11.025 = u × 0.75

Divide both side by 0.75

u = 11.025/0.75

u = 14.7 m/s

Therefore, the penny was travelling at 14.7 m/s before hitting the ground.

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A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 29.0 m/s. Then the truck trav
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Answer:

44.5s ; 22.64 m/s

Explanation:

The motion of the truck could be separated into 3 different phases :

First :

Time of motion :

Initial Velocity, u = 0 ; final velocity, v = 29 m/s

Acceleration, a = 2 m/s²

Recall: acceleration = change in velocity / time

Time = change in velocity / acceleration

Time = (29 - 0) / 2 = 14.5 second

Distance traveled = ((29 + 0) /2) * 14.5 = 210.25 m

Second :

Time = 25 seconds at constant speed

29 m/s for 25 seconds

v*t = 29 * 25 = 725 m

Third:

5 seconds before coming to rest

((29 + 0) /2) * 5

14.5 * 5 = 72.5 m

A.)

Length of journey = (14.5+ 25 + 5) = 44.5 seconds

B.)

Average velocity = total distance / total time taken

Average velocity = (210.25 + 725 + 72.5) / 44.5

= 1007.75 / 44.5

= 22.646067

= 22.64 m/s

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3 years ago
Why does the amount of water that flows in a river change during the year?
Nina [5.8K]
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4 years ago
The maximum heart rate during exercise for a 40 year old is
irina1246 [14]
<span>Target HR Zone 50-85% would be </span>90-153 beats per minuet. The average maximum heart rate 100% is 180 beats per minuet. 
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3 years ago
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

5 0
3 years ago
A uniform bar has two small balls glued to its ends. The bar has length L and mass M, while the balls each have mass m and can b
vazorg [7]

Answer:

Explanation:

Length of bar = L

mass of bar = M

mass of each ball = m

Moment of inertia of the bar about its centre perpendicular to its plane is

I_{1}=\frac{ML^{2}}{12}

Moment of inertia of the two small balls about the centre of the bar perpendicular to its plane is

I_{2}=2\times m\times \frac{L^{2}}{4}

I_{2}=\frac{mL^{2}}{2}

Total moment of inertia of the system about the centre of the bar perpendicular to its plane is

I = I1 + I2

I=\frac{ML^{2}}{12}+\frac{mL^{2}}{2}

I=\frac{(M +6m)L^{2}}{12}

8 0
3 years ago
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