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3 years ago

What is the velocity of a 30-kg box with a kinetic energy of 6,000 J? 64 m/s

Physics

1 answer:

Nikolay [14]3 years ago

8 0

Answer: 20 m/s

Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:

Ekinetic=(1/2)*(m*v^2)

then E=0.5*30Kg*(20 m/s)^2=15*400=6000J

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A 3.00 μF capacitor is charged to 480 V and a 4.00 μF capacitor is charged to 500 V . Part A These capacitors are then disconnec

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Q before connected = Q after connected C1V1+C2V2 = (C1+C2) V

C1= 3×10^-6 F

V1= 480v

C2= 4×10^-6 F

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(3×10^-6)×(480) + (4×10^-6)×(500) = (3×10^-6 + 4×10^-6) × V

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3 years ago

A cylindrical tank of methanol has a mass of 70 kg and a volume of 75 L. Determine the methanol’s weight, density, and specific

dem82 [27]

Answer:

Weight=686.7N, $\rho=933kg/m^{3}$ , S.G.=0.933, F=17.5N

Explanation:

So, the first value the problem is asking us for is the weight of methanol. (This is supposing there is a mass of methanol of 70kg inside the tank). We can find this by using the formula:

W=mg

so we can substitute the data the problem provided us with to get:

$W=70kg(9.81m/s^{2})$

which yields:

W=686.7N

Next, we need to find the density of methanol, which can be found by using the following formula:

$\rho=\frac{m}{V}$

we know the volume of methanol is 75L, so we can convert that to $m^{3}$ like this:

$75L*\frac{0.001m^{3}}{1L}=0.075m^{3}$

so we can now use the density formula to find our the methanol's density, so we get:

$\rho=\frac{m}{V}$

$\rho=\frac{70kg}{0.075m^{3}}$

$\rho=933.33kg/m^{3}$

Next, we can us these values to find the specific gravity of methanol by using the formula:

$S.G.=\frac{\rho_{sample}}{\rho_{H_{2}O}}$

when substituting the known values we get:

$S.G.=\frac{933.33kg/m^{3}}{1000kg/m^{3}}$

so:

S.G.=0.933

We can now find the force it takes to accelerate this tank linearly at $0.25m/s^{2}$

F=ma

$F=(70kg)(0.25m/s^{2})$

F=17.5N

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