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2 years ago

Please I need help ill give u 50 points

Physics

1 answer:

elena-s [515]2 years ago

3 0

Answer:

trasvere wave

Explanation:

first one

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS AND I NEED ALL CORRECT ANSWERS ONLY!!!

ryzh [129]

iron, cobalt and Nickel is the answer

6 0

2 years ago

1. A race car moves along a circular track of radius 100m at a velocity of 25m/s. (a) What is

goldenfox [79]

Given :

A race car moves along a circular track of radius 100m at a velocity of 25m/s.

To Find :

(a) What is the time taken to complete one lap of the circular track.

(b) What is the time taken for 10 laps.

Solution :

Circumference of circular track,

$C = 2\pi r\\\\C = 2\times \pi \times 100\ m\\\\C = 628.32 \ m$

a) Time taken to complete one lap is :

$t= \dfrac{628.32}{25}\ s\\\\t =25.13 \ s$

b) Time taken to complete 10 laps is :

$t_{10} = 10\times t\\\\t_{10} = 10\times 25.13\ s\\\\t_{10} = 251.3\ s$

Hence, this is the required solution.

6 0

2 years ago

What is the net power that a person with surface area of 1.20 m2 radiates if his emissivity is 0.895, his skin temperature is 27

marissa [1.9K]

Explanation:

Below is an attachment containing the solution.

4 0

2 years ago

A single conservative force acts on a 5.30-kg particle within a system due to its interaction with the rest of the system. The e

cupoosta [38]

Answer:

Given that

m = 5.3 kg

Fx = 2x + 4

We know that work done by force F given as

w= ∫ F. dx

Given that x=1.08 m to x=6.5 m

Fx = 2x + 4

w= ∫ F. dx

$w=\int_{1.08}^{6.5}(2x+4) .dx$

$w=\left [x^2+4x \right ]_{1.08}^{6.5}$

$w=(6.5^2-1.08^2)+4(6.5-1.08)\ J$

w=62.7 J

We know that potential energy given as

$F=-\dfrac{dU}{dx}$

∫ dU = -∫F.dx ( w= ∫ F. dx)

ΔU= -62.7 J

We know that form work power energy theorem

Net work = Change in kinetic energy

W= KE₂ - KE₁

62.7 =KE₂ - (1/2)x 5.3 x 3²

KE₂ = 86.55 J

This is the kinetic energy at 6.5m

8 0

2 years ago

Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human i

Varvara68 [4.7K]

Answer:

$0.0018833\ \text{m/s}$

Explanation:

$d$ = Distance of Andromeda Galaxy from Earth = $2.54\times 10^7\ \text{ly}$

$t$ = Time taken = $90\ \text{years}$

$c$ = Speed of light = $3\times 10^8\ \text{m/s}$

We have the relation

$t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}$

$c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}$

The required answer is $0.0018833\ \text{m/s}$ .

7 0

2 years ago