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taurus [48]
3 years ago
10

What is the mass, in grams, of a pure iron cube that has a volume of 3.70cm cubed?

Physics
1 answer:
Arlecino [84]3 years ago
7 0
To determine the amount in grams of the iron, we need data on the density of iron. From literature, it has a value of <span>p=7.9 g/cm3. We simply multiply the volume to the density. We do as follows:

mass = 3.70 (7.9) = 29.23 g Fe

Hope this answers the question. Have a nice day.</span>
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Answer:

A 1.0 min

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The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.

From the graph in the problem, we see that the initial mass of the isotope at time t=0 is

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The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:

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A 3. 5-a current is maintained in a simple circuit with a total resistance of 1500 ω. What net charge passes through any point i
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Electric Current:

Electric current is the flow of charge through a given circuit per unit time. Electric current is one of the components needed to calculate the electric power that a device needs to operate and do work. Electric current is measured in amperes (A), which is equal to:

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Electricity is produced when an electric current runs through a circuit.

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Who is responsible for documenting the crime scene in detail and collecting physical
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7 0
3 years ago
A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the
alex41 [277]

Answer:

(a)v_1 = a_1t_1 = 1.76 t_1

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1

(b) The velocity of the car before the driver begins braking is

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a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2

We can use the following equation of motion to calculate how far the car has travel since braking to stop

s_2 = v_1t_2 + a_2t_2^2/2

s_2 = 35.2*5 - 7.04*5^2/2 = 88 m

Also the distance from start to where the driver starts braking is

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So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

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