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almond37 [142]
3 years ago
12

A 150 g egg is dropped from 3.0 meters. The egg is moving at 4.4 m/s right before it hits the ground. The egg comes to a stop in

0.072 seconds.
What is the magnitude of force that the ground exerted on the egg?

0.66 N

9.2 N

13 N

180 N

Answer is 9.2 N
Physics
1 answer:
algol [13]3 years ago
8 0

Answer: Magnitude of the force exerted on the egg by the ground is 9.2N

Explanation:

Given the following :

Mass of egg (m) = 150g = 0.15kg

Height(h) from which egg is dropped = 3m

velocity of egg before hitting the ground (u) = 4.4m/s

Final velocity of egg (V) = 0

Time taken (t) = 0.072s

Magnitude of the force exerted on the egg by the ground can be found by applying Newton's 2nd law:

Momentum = mass × velocity

From Newton's second law:

Force = mass × change in Velocity with time ;

That is

F = m * ΔV / t

Inputting our values

F = 0.15 * (4.4 - 0) / 0.072

F = 0.15 × (4.4 / 0.072)

F = 0.15 × 61.11

F = 9.16N

F = 9.2N

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Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
Elena L [17]

Answer:

2.62898\times 10^{-6}\ C/m^3

1979.99974\ N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C

Volume charge density is given by

\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

8 0
3 years ago
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
sveticcg [70]

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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When the wave is produced, the points where the string is not moving are called nodes and where they are moving are called anti-nodes. The positions where nodes are produced, destructive interference occurs and where anti-nodes are produced, constructive interference occurs

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H_{2} +O_{2}=2(H_{2}O).

Resulting product is two molecules of water.

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