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1 year ago

Calculate the volume of material whose density is 0.798gcm-3 and a mass 25g

Physics

2 answers:

luda_lava [24]1 year ago

5 0

Explanation:

density equals mass/volume

Elena L [17]1 year ago

4 0

Answer:

$\mathrm{volume = \bf 31.33 \sapce\ cm^{-3}}$

Explanation:

To calculate the volume, we can use the following formula:

$\boxed{\mathrm{density = \frac{mass}{volume}}}$ .

In this case,

• density = 0.798 gcm⁻³

• mass = 25 g

Substituting values into the formula:

$\mathrm{0.798 \space\ gcm^{-3}= \frac{25 \space\ g}{volume}}$

⇒ $\mathrm{volume = \frac{25 \space\ g}{0.798 \space\ gcm^{-3}}}$

⇒ $\mathrm{volume = \bf 31.33 \sapce\ cm^{-3}}$

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Answer:

a) $P=2450\ Pa$

b) $\delta h=23.162\ cm$

Explanation:

Given:

height of water in one arm of the u-tube, $h_w=25\ cm=0.25\ m$

Gauge pressure at the water-mercury interface,:

$P=\rho_w.g.h_w$

we've the density of the water $=1000\ kg.m^{-3}$

$P=1000\times 9.8\times 0.25$

$P=2450\ Pa$

Now the same pressure is balanced by the mercury column in the other arm of the tube:

$\rho_w.g.h_w=\rho_m.g.h_m$

$1000\times 9.8\times 0.25=13600\times 9.8\times h_m$

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Now the difference in the column is :

$\delta h=h_w-h_m$

$\delta h=25-1.838$

$\delta h=23.162\ cm$

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2 years ago

The relationship between sharks and remoras is characterized TRUE OR FALSE

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You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle

AlekseyPX

Answer:

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

Explanation:

When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that

Work done by all the forces = change in kinetic energy of the system

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here we know that

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also we know that the length of the incline is given as

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now we have

$- mgh - \mu_k mgcos\theta(\frac{h}{sin\theta}) = -\frac{1}{2}mv^2$

so we have

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2 years ago

Sam, whose mass is 75 kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 160 N and a coefficien

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Answer:

top speed = 17.25

Total height = 281.19 m

Explanation:

given data

mass = 75 kg

thrust = 160 N

coefficient of kinetic friction = 0.1

solution

we get here frictional force acting that is

frictional force = $\mu *m*g$ .............1

frictional force = 0.1 × 75 × 9.8

frictional force = 73.5 N

and

Net force acting will be F = 160 - 73.5 N

F = 86.5 N

Acceleration in the First 15 second will be

F = ma .........2

86.5 = 75 × a

a = 1.15 m/s²

and

now After 15 second the velocity will be as

v = u + at ..........3

here u is 0

so v will be

V = 1.15 × 15

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and

now we get travels distance S in 15 s

s = u × t + 0.5 × a × t²

here u is 0

so distance s will be

s = 0.5 × a × t²

s = 0.5 × 1.15 × 15²

s = 129.37 m

and

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F = $\mu *m*g$

m a = $\mu *m*g$

a = $\mu* g$

a = - 0.98

here it is negative it mean downward nature of acceleration

and

now we get distance s by this formula

V² - u² = 2 a s

here v velocity is 0 and

u initial velocity is 17.25 m/s

put here value

0 - 17.25² = 2 × (-0.98) × s

solve it we get

s = 151.82 m

Total height is

Total height = 129.37 m + 151.82 m

Total height = 281.19 m

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