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3 years ago

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Kevin goes bowling. Whenever he bowls the ball, he transfers energy from his hand to the bowling ball. The amount of energy befo

re the transfer is ____________ the amount of energy after the transfer.
A. Less than
B. More than
C. Equal to
D. Not related to

1 answer:

g100num [7]3 years ago

3 0

Answer:the answer is C

Explanation:

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An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?

Serjik [45]

The image of the object is 8cm to the left of the lens (D)

<h3></h3>

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

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3 years ago

Read 2 more answers

A standing wave pattern is created on a string with mass density μ = 3.4 × 10-4 kg/m. A wave generator with frequency f = 61 Hz

uranmaximum [27]

Answer:

1) λ = 0.413 m , 2)v = 25,213 m / s , 3) T = 0.216 N , 4) m = 22.04 10-3 kg

Explanation:

1) The resonance occurs when the traveling wave bounces at the ends and the two waves are added, the ends as they are fixed have a node, the wavelength and the length of the string are related

λ = 2L / n n = 1, 2, 3 ...

In this case L = 0.62 m and n = 3

Let's calculate

λ = 2 0.62 / 3

λ = 0.413 m

2) the velocity related to wavelength and frequency

v = λ f

v = 0.413 61

v = 25,213 m / s

3) let's use the equation

v = √T /μ

T = v² μ

T = 25,213² 3.4 10⁻⁴

T = 0.216 N

4) the rope tension is proportional to the hanging weight

T-W = 0

T = W

W = m g

m = W / g

m = 0.216 / 9.8

m = 22.04 10-3 kg

5) n = 2

λ = 2 0.62 / 2

λ = 0.62 m

6) v = λ f

v = 0.62 61

v = 37.82 m / s

7) T = v² μ

T = 37.82² 3.4 10⁻⁴

T = 0.486 N

8) m = W / g

m = 0.486 / 9.8

m = 49.62 10⁻³ kg

9) n = 1

λ = 2 0.62

λ = 1.24 m

v = 1.24 61

v = 75.64 m / s

T = v² miu

T = 75.64² 3.4 10⁻⁴

T = 2.572 10⁻² N

m = 2.572 10⁻² / 9.8

m = 262.4 10⁻³ kg

5 0

3 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c

Travka [436]

Answer:

22.1 V

Explanation:

We are given that

$A_1=0.7 cm^2=0.7\times 10^{-4} m^2$

$A_2=2.5 cm^2=2.5\times 10^{-4} m^2$

$A_3=2.2 cm^2=2.2\times 10^{-4} m^2$

$A_4=3 cm^2=3\times 10^{-4} m^2$

Using $1cm^2=10^{-4} m^2$

We know that

$R=\frac{\rho l}{A}$

In series

$R=R_1+R_2+R_3+R_4$

$R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}$

$R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}$

Substitute the values

$R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})$

$R=\rho l(2.62\times 10^4)$

$V=145 V$

$I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}$

Voltage across the 2.5 square cm wire= $IR=I\times \frac{\rho l}{A_2}$

Voltage across the 2.5 square cm wire= $\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V$

Voltage across the 2.5 square cm wire=22.1 V

6 0

3 years ago