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2 years ago

What is the density of a block of wood with a mass of 12g and a volume of 6 cm3?

Physics

1 answer:

scZoUnD [109]2 years ago

5 0

<h2>Answer: 2 g/cm³</h2>

Explanation:

Density = mass ÷ volume

= 12 g ÷ 6 cm³

= 2 g/cm³

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3 years ago

You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in

FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, $V=108\ cm^3$

We know that the volume of the cylinder is given by :

$V=\pi r^2 h$

$108=\pi r^2 h$

$h=\dfrac{108}{\pi r^2}$ ............(1)

The upper area is given by :

$A=32r^2+2\pi rh$

$A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}$

$A=32r^2+\dfrac{216}{r}$

For maximum area, differentiate above equation wrt r such that, we get :

$\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}$

$64r-\dfrac{216}{r^2}=0$

$r^3=\dfrac{216}{64}$

r = 1.83 m

Dividing equation (1) with r such that,

$\dfrac{h}{r}=\dfrac{108}{\pi r}$

$\dfrac{h}{r}=\dfrac{108}{\pi 1.83}$

$\dfrac{h}{r}=59 \pi$

Hence, this is the required solution.

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2 years ago

The manufacturer of a bulletproof vest wants the vest to be able to stop a bullet with a mass of 0.4 kg and a velocity of 1800 m

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3 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

3 0

3 years ago