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2 years ago

Question 2 of 5

Physics

1 answer:

Setler79 [48]2 years ago

5 0

B.when it is leaving the bat

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Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a

____ [38]

Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, $v=2.79\times 10^8\ m/s$

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

$t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s$

Time period of oscillation measured by the observer is :

$t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s$

So, the time period of oscillation measured by the observer is 5.79 seconds.

3 0

3 years ago

The initial volume reading in a graduated cylinder is 30 mL. The mass of an irregular shape of an unknown metal piece is 55.3 g.

Vadim26 [7]

Answer:

7.9 $\frac{gr}{cm^3}$

Explanation:

When we put the metal piece in the liquid (which is in the graduated cylinder), how much it goes up is equal to the volume of the piece we inserted.

So now we know that the volume of that piece of unknown metal is 7mL (which is the same as 7 $cm^3$ ).

Density is $\frac{mass}{volume}$ .

So the density of that piece of metal is $\frac{55.3g}{7cm^3}$

Which leaves us with a final density of 7.9 $\frac{gr}{cm^3}$

6 0

3 years ago

How do u do 5? Show me plz

nikitadnepr [17]

Solve the equation.it is hard to see

6 0

2 years ago

Which of the following statements is true?

hram777 [196]

All the radiation from stars is the result of nuclear fusion in their cores.

4 0

3 years ago

Read 2 more answers

A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____

Rama09 [41]

Answer:

Magnification, m = -0.42

Explanation:

It is given that,

Height of diamond ring, h = 1.5 cm

Object distance, u = -20 cm

Radius of curvature of concave mirror, R = 30 cm

Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)

Using mirror's formula :

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$ , f = focal length of the mirror

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{-20}$

v = -8.57 cm

The magnification of a mirror is given by,

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-8.57)}{-20}$

m = -0.42

So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.

5 0

3 years ago