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3 years ago

8.

Physics

1 answer:

vladimir1956 [14]3 years ago

3 0

Answer:

a. petrification

Explanation:

tar seeps = natural trade that, because of its close proximity to the ground surface, seeps from the cracks in the Earth or between rocks forming pits or pools (tar pits)

amber = fossilized resin produced by extinct coniferous trees, typically yellow in color

mummification = a process in which the skin and flesh of a corpse can be preserved by embalming and drying

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An object is 30 cm in front of a converging lens with a focal length of 10 cm. Use ray tracing to determine the location of the

Aleksandr-060686 [28]

Answer:

i). Inverted

ii). Magnification of the image = -0.5

iii). Real

Explanation:

As shown in the ray diagram attached,

An object AB has been placed in front of converging lens (convex lens) at u = 30 cm.

F (Focus) of the lens is at 10 cm. So F = 10 cm

By analyzing the ray diagram we can measure the distance of the image on the other side of the lens (By counting the small blocks of the graph)

V = 15 cm

Characteristics of the image is:

i) Inverted

ii) Magnification of the image = $-\frac{v}{u}=-\frac{15}{30}$

= -0.5

ii) Real

4 0

3 years ago

A wall clock has a minute hand with a length of 0.55 m and an hour hand with a length of 0.26 m. Take the center of the clock as

Lemur [1.5K]

Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

Explanation:

Given that,

Length of minute hand = 0.55 m

Length of hour hand = 0.26 m

The time taken by the minute hand to complete one revelation is

$T= 3600\ sec$

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\dfrac{2\pi}{T}$

Put the value into the formula

$\omega=\dfrac{2\pi}{3600}$

$\omega=0.001745\ rad/s$

We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock

Using formula of acceleration

$a=r\omega^2$

Put the value into the formula

$a=0.55\times(0.001745)^2$

$a=1.675\times10^{-6}\ m/s^2$

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

3 0

3 years ago

The equation of a transverse wave traveling along a very long string is y 6.0 sin(0.020px 4.0pt), where x and y are expressed in

zhannawk [14.2K]

Answer:

given

y=6.0sin(0.020px + 4.0pt)

the general wave equation moving in the positive directionis

y(x,t) = ymsin(kx -?t)

a) the amplitude is

ym = 6.0cm

we have the angular wave number as

k = 2p /?

? = 2p / 0.020p

=1.0*102cm

the frequency is

f = ?/2p

= 4p/2p

= 2.0 Hz

the wave speed is

v = f?

= (100cm)(2.0Hz)

= 2.0*102cm/s

since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction

the maximum transverse speed is

umax =2pfym

= 2p(2.0Hz)(6.0cm)

= 75cm/s

we have

y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]

= -2.0cm

6 0

3 years ago

How to do this question

mestny [16]

thanks again and have to go to the store and get some rest I will be there at puno my phone is not working and I have a few questions about the week and I have a few questions about the week and I have a few questions about the week and I have a few questions about the week and I have a few questions about the week and I have a few questions about the week and I have a few questions about the week and I have a few questions about the week and I have a few questions about the week and I have a few questions about you

4 0

2 years ago

Calculate the maximal friction force for a parked car between the rubber tires and a wet street. Assume the car’s mass is 1600 k

NikAS [45]

Answer:

$Fr=12544N$

Explanation:

1. Find the equation of eht maximal friction force:

The maximal friction force is given by the equation $Fr=usmg$ , where μs is the static friction coefficient, m is the car´s mass and g is the gravitational force.

2. Replace values in the equation to find the answer:

$Fr=0.8*1600kg*9.8\frac{m}{s^{2}}$

$Fr=12544N$

5 0

3 years ago