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2 years ago

11

How much heat is required to evaporate 0.15 kg of lead at 1750°C, the boiling point for lead? The heat of vaporization for lead

is Lv = 871 × 103 J/kg.

1 answer:

lara [203]2 years ago

3 0

Answer:

Heat required = mass× latent heat Q = 0.15 × 871 ×

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What is the kinetic energy of a 150 kg bear running at 3 m/s?

OverLord2011 [107]

Explanation:

KE = ½ mv²

KE = ½ (150 kg) (3 m/s)²

KE = 675 J

5 0

3 years ago

Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th

lapo4ka [179]

Answer:

$T_f=5.0116^{\circ}C$

Explanation:

Given:

mass of water, $m_w=0.25\ kg$

initial temperature of water, $T_i_w=20^{\circ}C$

initial temperature of pan, $T_i_p=173^{\circ}C$

mass of pan, $m_p=0.6\ kg$

mass of water evapourated, $m_v=0.03\ kg$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

specific heat of aluminium pan, $c_a=900\ J.kg^{-1}.K^{-1}$

latent heat of vapourization, $L=2256000\ J.kg^{-1}$

<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

$m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})$

$0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)$

$T_f=5.0116^{\circ}C$

5 0

3 years ago

B c<br>_ - _<br>a b<br>( a=3 b=-5 c=6)

Pachacha [2.7K]

(-5)/3 - 6/(-5)
You can solve it now :)

7 0

2 years ago

A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc

german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = $\frac{1}{2} kx^2$ and the kinetic energy gained by the box is given by KE = $\frac{1}{2} mv^2$

Since work done by the spring = kinetic energy gained by the box we have

$\frac{1}{2} mv^2$ = $\frac{1}{2} kx^2$ therefore we have v = $\sqrt{\frac{kx^2}{m} }$ = $x\sqrt{\frac{k}{m} }$ = $0.15\sqrt{\frac{300}{0.2} }$ = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = $\frac{v^2}{2g}$ = $\frac{5.81^{2} }{2*9.81}$ = 1.72 m

6 0

3 years ago

A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a

ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω / t

(ω = √g/R)

∝ = $\frac{1}{t } \sqrt{\frac{g}{R} } }$

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = $\frac{2FR}{MR^2 + 2mr^2}$

Now we have ,

∝ = $\frac{1}{t} \sqrt{\frac{g}{R} }$ , ∝ = $\frac{2FR}{MR^2+2mr^2}$

equating both values

We have,

$F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }$

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

$F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N$

Each thruster has to applied a force of 294.5N in tangential direction

3 0

3 years ago