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3 years ago

A standard 1 kilogram weight is a cylinder 41.5 mm in height and 44.0 mm in diameter. what is the density of the material?

Physics

1 answer:

Korvikt [17]3 years ago

6 0

Answer;

=15855.40 kg/m^3

Explanation;

Volume (V) of the cylinder = pi x r^2 x h

V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3

V = 6.307 x 10^-5 m^3

By density = m/V

mass = 1 kg

density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3

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A bearing if an angle is measured clockwise from north direction.

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2 years ago

A block of mass M is suspended from two identical springs of negligible mass, spring constant k, and unstretched length L. First

34kurt

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The answer is (C)X1=2X2

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3 years ago

A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c

Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current density $J=2.8\times10^{7}\ A/m^2$

We need to calculate the current

Using formula of current density

$J = \dfrac{I}{A}$

$I=J\timesA$

Where, J = current density

A = area

I = current

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$I=1.97=2.0\ A$

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7 0

3 years ago

Several light bulbs, each of resistance 1.5 Ω, are connected in a series across a 120 V source of emf. If the current through th

Leni [432]

<h3><u>Answer;</u></h3>

40 light bulbs

<h3><u>Explanation</u>;</h3>

The total resistance of components or bulbs in series is given as the sum of resistance of all the components.

Thus; if there are bulbs in series each with a resistance of 1.5 Ω, the the total resistance will be; 1.5nΩ

From the ohms law;

V = IR , where V is the voltage, I is the current and R is the resistor.

Thus; R = V/i

R = 120/2

= 60 Ω

But, there are n bulbs each with 1.5 Ω; thus there are;

n = 60/1.5

<u> = 40 Bulbs </u>

7 0

3 years ago

The coefficients of friction between the load and the flatbed trailershown are μs = 0.40 and μk = 0.30. Knowing that the speed o

SOVA2 [1]

Answer:

50.97 m

Explanation:

m = Mass of truck

$\mu_s$ = Coefficient of static friction = 0.4

v = Final velocity = 0

u = Initial velocity = 72 km/h = $\dfrac{72}{3.6}=20\ \text{m/s}$

s = Displacement

Force applied

$F=ma$

Frictional force

$f=\mu_s mg$

Now these forces act opposite to each other so are equal. This is valid for the case when the load does not slide

$ma=\mu_s mg\\\Rightarrow a=\mu_s g\\\Rightarrow a=0.4\times 9.81\\\Rightarrow a=3.924\ \text{m/s}^2$

Since the obect will be decelerating the acceleration will be $-3.924\ \text{m/s}^2$

From the kinematic equations we have

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -3.924}\\\Rightarrow s=50.97\ \text{m}$

So, the minimum distance at which the car will stop without making the load shift is 50.97 m.

5 0

3 years ago