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Alex73 [517]
3 years ago
13

Can someone plz help me DUE TOMORROW

Physics
1 answer:
aksik [14]3 years ago
3 0
Hey the answer to the question is
m = 0.40
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Which description is evidence of chemical weathering?
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b.

Explanation: I do not know much about this but the answer that i think it is was b.

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Find the frequency of a wave of wavelength 2.5m and speed 400 m/s
Sedaia [141]

Answer:

The frequency of wave is 160Hz.

Explanation:

Given that the formula of speed is V = f×λ where V represents speed, f is frequency and λ is wavelength.

So first thing, you have to make frequency the subject by dividing wavelength on both sides :

v = f \times λ \:

v \div  λ =  f \times λ \div λ

f =  \frac{v}{λ}

Next you have to substitute the value of v and f into the formula :

Let λ = 2.5m,

Let v = 400m/s,

f =  \frac{400}{2.5}

f = 160Hz

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3 years ago
When an object is heated, its molecules
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A

Explanation:

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3 years ago
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A car with a constant velocity of 22 m/s is driven for 6.8 s. How far did it travel? (v =∆ d/∆t)
Genrish500 [490]

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3 years ago
The frictionless system shown is released from rest. After the right-hand mass has risen 75 cm, the object of mass 0.50m falls l
Andreas93 [3]

Let a be the acceleration of the masses. By Newton's second law, we have

• for the masses on the left,

1.3mg - T = 1.3ma

where T is the magnitude of tension in the pulley cord, and

• for the mass on the right,

T - mg = ma

Eliminate T to get

(1.3mg - T) + (T - mg) = 1.3ma + ma

0.3mg = 2.3ma

\implies a = \dfrac{0.3}{2.3}g \approx 0.13g \approx 1.3 \dfrac{\rm m}{\mathrm s^2}


Starting from rest and accelerating uniformly, the right-hand mass moves up 75 cm = 0.75 m and attains an upward velocity v such that

v^2 = 2a(0.75\,\mathrm m) \\\\ \implies v \approx \sqrt{2\left(1.3\frac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx 1.4\dfrac{\rm m}{\rm s}

When the 0.5m mass is released, the new net force equations change to

• for the mass on the right,

mg - T' = ma'

where T' and a' are still tension and acceleration, but not having the same magnitude as before the mass was removed; and

• for the mass on the left,

T' - 0.8mg = 0.8ma'

Eliminate T'.

(mg - T') + (T' - 0.8mg) = ma' + 0.8ma'

0.2mg = 1.8 ma'

\implies a' = \dfrac{0.2}{1.8}g = \dfrac19 g \approx 1.1\dfrac{\rm m}{\mathrm s^2}

Now, the right-hand mass has an initial upward velocity of v, but we're now treating down as the positive direction. As it returns to its starting position, its speed v' at that point is such that

{v'}^2 - v^2 = 2a'(0.75\,\mathrm m) \\\\ \implies v' \approx \sqrt{\left(1.4\dfrac{\rm m}{\rm s}\right)^2 + 2\left(1.1\dfrac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx \boxed{1.9\dfrac{\rm m}{\rm s}}

3 0
2 years ago
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