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fiasKO [112]
3 years ago
10

A 0.660 kg ball is dropped from rest from the top of a building and falls for 5.65 seconds. How tall was the building ?

Physics
1 answer:
kirill115 [55]3 years ago
6 0

Answer:

The height of the building is approximately 156.58 m

Explanation:

The mass of the ball dropped from rest from the building top = 0.660 kg

The time in which the ball falls, t = 5.65 seconds

The height, h, of the building is given from the following equation of motion;

h = u·t + ¹/₂·g·t²

Where;

u = The initial velocity of the ball = 0 m/s

g = The acceleration due to gravity = 9.81 m/s²

Plugging in the values, we have;

h = 0 × 5.65 + ¹/₂ × 9.81 × 5.65² ≈ 156.58 m

The height of the building, h ≈ 156.58 m.

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Answer:

Explanation:

A car travels 6.0 km to the north and then 8.0 km to the east. The intensity of the position vector, in relation to the starting point is: a) 14 km b) 2.0 km c) 12 km d) 10 km e) 8.0 km

Check attachment for diagram

The intensity of the position vector is equal to the displacement,

So, to calculate the displacement, we need to find the length of the straight line from starting point to end point.

So, applying Pythagorean theorem

c² = a² + b²

R² = 6² + 36²

R² = 36 + 64

R² = 100

R = √100

R = 10 km.

Verifique el adjunto para ver el diagrama

La intensidad del vector de posición es igual al desplazamiento,

Entonces, para calcular el desplazamiento, necesitamos encontrar la longitud de la línea recta desde el punto inicial hasta el punto final.

Entonces, aplicando el teorema de Pitágoras

c² = a² + b²

R² = 6² + 36²

R² = 36 + 64

R² = 100

R = √100

R = 10 km.

7 0
3 years ago
Only about 10% of the electromagnetic energy from an incandescent lightbulb is visible light. The bulb radiates most of its ener
AlekseyPX

Answer:

<em>1. 1.48 W/</em>m^{2}<em></em>

<em>2. 4.93 x </em>10^{-9}  J/m^{3}<em></em>

<em>3. 33.3 V/m</em>

<em>4. 1.11 x </em>10^{-7} T<em></em>

Explanation:

The step by step explanation to the question is contained in the attached images;

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A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
belka [17]

Answer:

a)n= 3.125 x 10^{19 electrons.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

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Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x 10^{-3 m

radius 'r' = d/2 => 1.025 x 10^{-3 m

no. of electrons 'n'= 8.5 x 10^{28}

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I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrons.

b)  the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) The typical speed'V_{d' of an electron is given by:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) According to these equations,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

 

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