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3 years ago

A certain material has a mass of 565 g while occupying 50 cm3 of space. What is this material?

Physics

1 answer:

EastWind [94]3 years ago

8 0

<u>Answer:</u>

Lead

<u>Explanation:</u>

To get the density of the material, the formula would be:

mass divided by volume which is given by $d = \frac { m } { v }$ .

Here in this problem, we are given a mass of $565 g$ which occupies a volume of $50 cm^3$ .

So plugging the data in the above formula to find the density:

Density = $\frac { 565 } { 50 } = 11.3$

From the table, we can see that the material is Lead which has a density of 11.3c/cm^3.

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Two protons (each with q = 1.60 x 10-19)

otez555 [7]

Answer:

230.4 N

Explanation:

From the question given above, the following data were obtained:

Charge (q) of each protons = 1.6×10¯¹⁹ C

Distance apart (r) = 1×10¯¹⁵ m

Force (F) =?

NOTE: Electric constant (K) = 9×10⁹ Nm²/C²

The force exerted can be obtained as follow:

F = Kq₁q₂ / r²

F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²

F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰

F = 2.304×10¯²⁸ / 1×10¯³⁰

F = 230.4 N

Therefore, the force exerted is 230.4 N

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2 years ago

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Marrrta [24]

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2 years ago

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Convert 5.7 miles to km

gayaneshka [121]

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3 years ago

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Define second class lever

Aleks04 [339]

Answer:

Please find detailed explanation of second class levers below

Explanation:

Levers are one of the classes of machine that possesses three levels namely: first class, second class and third claas. A second class lever is the level of levers in which the load (L) is in between the pivot (F) and the effort (E).

Examples of second class levers include; wheelbarrow, a bottle opener etc. In the bottle opener for example, the bottle lid (load) is in between the pivot of the opener and the hand opening it (effort).

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2 years ago

an 1150kg elevator moving down speeds up at a rate of 3.5m/s. what is the tension in the supporting cables?

gtnhenbr [62]

Answer:

The tension force in the supporting cables is 7245N

Explanation:

There are two forces acting on the elevator: the force of gravity pointing down (+) with magnitude (elevator mass) x (gravitational acceleration), and the tension force of the cable pointing up (-) with an unknown magnitude F. The net force is the sum of these forces:

$F_{net} = F_g - F = m\cdot g - F\\$

We are given the resulting acceleration along with the mass, i.e., we know the net force, allowing us to solve for F:

$1150kg\cdot 3.5\frac{m}{s^2}= 1150kg \cdot 9.8\frac{m}{s^2}-F\\\implies F = 1150kg\cdot(9.8-3.5)\frac{m}{s^2}= 7245N$

The tension force F in the supporting cables is 7245N

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3 years ago