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2 years ago

Can anyone plz help me

Physics

1 answer:

Alekssandra [29.7K]2 years ago

3 0

Answer:

1.) volume = 8 cm^3

2.) density = 2.6 g/cm^3

Explanation:

1.) volume =

$side {}^{3}$

=》

${2}^{3}$

=》

$8 \: cm {}^{3}$

2.) density =

$\frac{mass}{volume}$

=》

$\frac{20.8}{8}$

=》

$2.6 \: g \: cm {}^{ - 3}$

density = 2.6

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Answer:

= 0.086J

Explanation:

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sides of triangle = 0.800m

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U = 9.0 × 10⁹ ((3 * 1.60 × 10⁻⁶)² / (0.800))

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3 years ago

A 50.0 kg object is moving at 18.2 m/s when a 200 N force is applied opposite the direction of the objects motion, causing it to

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Answer:

t = 1.4[s]

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

$P=m*v\\or\\P=F*t$

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 50 [kg]

v = velocity [m/s]

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t = time = [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

$(m_{1}*v_{1})-F*t=(m_{1}*v_{2})$

where:

m₁ = mass of the object = 50 [kg]

v₁ = velocity of the object before the impulse = 18.2 [m/s]

v₂ = velocity of the object after the impulse = 12.6 [m/s]

$(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]$

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3 years ago

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The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha

saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0

3 years ago

Can anyone plz help me ​

Can anyone plz help me