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3 years ago

10

A cup of coffee is considered a Blank Space __________.

1 answer:

Sauron [17]3 years ago

4 0

When you brew a coffee you're creating a Solution.

Tiny particles from coffee grounds, or solutes, are dissolved in water, the solvent.

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a 15kg television sits on a shelf at a height of 0.3 m how much gravitational potential energy is added to the television when i

DedPeter [7]

Answer:

<h2>103 Joules</h2>

Explanation:

In this problem we are required to find the potential energy possessed by the television

Given data

mass of television m = 15 kg

height added above the ground, h= 1-0.3 = 0.7 m

acceleration due to gravity g = 9.81 m/s^2

apply the formula for potential energy we have

P.E= m*g*h

P.E = 15*9.81*0.7 = 103 Joules

7 0

3 years ago

A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal in

erica [24]

Explanation:

Given that,

Mass of a freight car, $m_1=30,000-kg$

Speed of a freight car, $u_1=0.85\ m/s$

Mass of a scrap metal, $m_2=110,000\ kg$

(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :

$30000\times 0.85=(30000+110000)V\\\\V=\dfrac{30000\times 0.85}{30000+110000}\\\\=0.182\ m/s$

So, the final velocity of the loaded freight car is 0.182 m/s.

(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy

$\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J$

Lost in kinetic energy is 8518.82. Negative sign shows loss.

6 0

2 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

3 years ago

A hollow sphere of inner radius 8.82 cm and outer radius 9.91 cm floats half-submerged in a liquid of density 948.00 kg/m^3. (a)

kari74 [83]

Answer:

a) 0.568 kg

b) 474 kg/m³

Explanation:

Given:

Inner radius = 8.82 cm = 0.0882 m

Outer radius = 9.91 cm = 0.0991 m

Density of the liquid = 948.00 Kg/m³

a) The volume of the sphere = $\frac{4\pi}{3}\times(0.0991^2-0.0882^2)$

or

volume of sphere = 0.0012 m³

also, volume of half sphere = $\frac{\textup{Total volume}}{\textup{2}}$

or

volume of half sphere = $\frac{\textup{0.0012}}{\textup{2}}$

or

Volume of half sphere =0.0006 m³

Now, from the Archimedes principle

Mass of the sphere = Weight of the volume of object submerged

or

Mass of the sphere = 0.0006× 948.00 = 0.568 kg

b) Now, density = $\frac{\textup{Mass}}{\textup{Volume}}$

or

Density = $\frac{\textup{0.568}}{\textup{0.0012}}$

or

Density = 474 kg/m³

8 0

3 years ago

When we perform an experiment we

Verizon [17]

Write a:
● Hypothesis
●Procedures
●Materials
●Data Table
●Conclusion
●Variables (Independent and dependent)
●Bibliography

4 0

3 years ago