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riadik2000 [5.3K]
3 years ago
10

A meteor is approaching Earth. Which statement about its motion is true?

Physics
2 answers:
Aleksandr-060686 [28]3 years ago
6 0
<h2>Answer:</h2>

 。・:*:・゚★,。・:*:・゚☆여보세요。・:*:・゚★,。・:*:・゚☆ 

꘎♡━━━━━━━━━━━━━━━♡꘎

⋆ ˚。⋆୨୧˚ The answer is below  ˚୨୧⋆。˚ ⋆

꘎♡━━━━━━━━━━━━━━━♡꘎

B) The meteor accelerates. It accelerates due to Earths gravitational pull on the meteor.

☆♬○♩●♪Have a nice day♩✧♪●♩○♬☆

:D

Maslowich3 years ago
3 0
B. The meteor accelerates
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A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
ZanzabumX [31]

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

8 0
3 years ago
A special electronic sensor is embedded in the seat of a car thattakes riders around a circular loop-the-loop at an amusement pa
Igoryamba

Answer:

v = 17.9 m/s

Explanation:

As we know that the normal force measured by the sensor before the ride is started is given as

F_n = mg = 990 N

now when the rider has reached at the top position of the loop then the normal force is given as

F_n' = 360 N

now at the top position we have

F_n' + mg = ma

F_n' + mg = \frac{mv^2}{R}

so we have

990 + 360 = \frac{990 v^2}{9.8 \times 24}

v = 17.9 m/s

6 0
3 years ago
Read 2 more answers
How does the atmosphere interact with the geosphere
Mama L [17]

Answer:

Explanation:

The atmosphere is the gaseous portion of the earth. It consists of different molecules of gas.

The geosphere is the solid portion of the earth which include the crusts, mantle and the core.

Earth is a dynamic planet. Our planet is dynamic in the sense that it is constantly changing and all its parts interacts with one another.

The gases in the atmosphere such a CO₂, H₂O, Nitrogen oxides originates from volcanic processes from deep within the earth. Hydrothermal vents and black smokers constantly release gases into the atmosphere.

The atmosphere plays a lot of roles in determining weather and climatic conditions. Agents of denudation like wind, water and glaciers are connected to the movement of gaseous portion of the earth. As new rocks forms on the crust, wind, water and glaciers acts on them. This process plays a central role in the rock cycle. The rock cycle would not be complete without agents of denudation which are strongly connected to the workings of atmospheric gases and materials like dusts.  

Therefore we see that the geosphere and atmosphere are linked.

4 0
3 years ago
1). An Owl and bat share the same kingdom and phylum; an owl and a cardinal share the same kingdom, phylum, and class. The owl a
kolbaska11 [484]
1. cardinal

2.  Protista

Glad to help :)
7 0
3 years ago
Read 2 more answers
A person with a mass of 15kg is walking uphill at a velocity of2m/s. What is the walker's momentum?
Nadusha1986 [10]

Answer:

30 kgm/s

Explanation:

The momentum of a body is given by the product of mass and  its velocity.

∴ momentum = mv

                        = 15 × 2

                        = 30 kgm/s

4 0
3 years ago
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