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2 years ago

15

A block slides down a rough ramp with a 30-degree incline as shown.

2 answers:

Studentka2010 [4]2 years ago

6 0

Answer: The first one

Explanation: My best guess is in the pic attached. Hope this helps.

Katen [24]2 years ago

5 0

Answer:

12 N to the right

Explanation:

EDGE 2021

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The behavior of which objects are better explained by Einstein’s general theory of relativity than Newton’s universal law of gra

liq [111]

If I had to go with any of those answers, It would be A maybe D, But im not too sure on how to decide between them. Because Einstein mentioned the sun in his theory which has a very large mass <span> 1.989 x 10 with a exponent of 30 to be exact. Hope this helped though.</span>

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2 years ago

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5kg of copper ball and I kg of water are at the same temperature of 31 C. Which

blagie [28]

Answer:

r

Explanation:

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3 years ago

WARRIOR [948]

Answer:

Explanation:distance-time =speed

a,V =s/t

V=2m/2s

V=1m/s

b,v=s/t

V=80m/40s

V= 2m/s

The average speed is 2m/s

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1 year ago

A string of mass 60.0 g and length 2.0 m is fixed at both ends and with 500 N in tension. a. If a wave is sent along this string

Darya [45]

Answer:

a

The speed of wave is $v_1 = 129.1 \ m/s$

b

The new speed of the two waves is $v = 129.1 \ m/s$

Explanation:

From the question we are told that

The mass of the string is $m = 60 \ g = 60 *10^{-3} \ kg$

The length is $l = 2.0 \ m$

The tension is $T = 500 \ N$

Now the velocity of the first wave is mathematically represented as

$v_1 = \sqrt{ \frac{T}{\mu} }$

Where $\mu$ is the linear density which is mathematically represented as

$\mu = \frac{m}{l}$

substituting values

$\mu = \frac{ 60 *10^{-3}}{2.0 }$

$\mu = 0.03\ kg/m$

So

$v_1 = \sqrt{ \frac{500}{0.03} }$

$v_1 = 129.1 \ m/s$

Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )

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3 years ago

A 2.10 cm × 2.10 cm square loop of wire with resistance 1.30×10−2 Ω has one edge parallel to a long straight wire. The near edge

Troyanec [42]

Answer:

$I_{l} =44.84 \mu A$

Explanation:

given,

side of square loop = a = 2.10 cm

Resistance of the wire = 1.30×10⁻² Ω

Length of the loop = c = 1.10 cm

rate of increasing current = 130 A/s

$\phi = \dfrac{\mu_0Ib}{2\pi}(ln(\dfrac{c+a}{c}))$

$\dfrac{d\phi}{dt}= \dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))$

$I_{l} = \dfrac{V}{R}$

$I_{l} = \dfrac{1}{R}\dfrac{d\phi}{dt}$

$I_{l} = \dfrac{1}{R}\dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))$

$I_{l} = \dfrac{1}{1.3 \times 10^{-2}}\dfrac{4\pi\times 10^{-7}\times 0.021}{2\pi}\times 130\times (ln(\dfrac{0.011+0.021}{0.011}))$

$I_{l} =44.84 \times 10^{-6}\A$

$I_{l} =44.84 \mu A$

3 0

3 years ago