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Savatey [412]
3 years ago
14

a mass of 1.00 kg of water at temperature T is poured from a height of 0.100 km into a vessel containing water of the same tempe

rature T, and a temperature change of 0.100 is measured. What mass of water was in the vessel? ignore heat flow into the vessel, the thermometer etc.
Physics
1 answer:
Mariana [72]3 years ago
6 0

Answer:

1.34352 kg

Explanation:

m_w = Mass of water falling = 1 kg

h = Height of fall = 0.1 km

\Delta T = Change in temperature = 0.1

c = Specific heat of water = 4186 J/kg K

g = Acceleration due to gravity = 9.81 m/s²

m_v = Mass of water in the vessel

Here the potential energy will balance the internal energy

m_wgh=m_wc\Delta T+m_vc\Delta T\\\Rightarrow m_v=\dfrac{m_wgh-m_wc\Delta T}{c\Delta T}\\\Rightarrow m_v=\dfrac{m_wgh}{c\Delta T}-m_w\\\Rightarrow m_v=\dfrac{1\times 9.81\times 100}{4186\times 0.1}-1\\\Rightarrow m_v=1.34352\ kg

Mass of the water in the vessel is 1.34352 kg

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Who what where when why how ?
6 0
3 years ago
A rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 kg
anyanavicka [17]

Answer:

0.4113772 s

Explanation:

Given the following :

Mass of bullet (m1) = 8g = 0.008kg

Initial horizontal Velocity (u1) = 280m/s

Mass of block (m2) = 0.992kg

Maxumum distance (x) = 15cm = 0.15m

Recall;

Period (T) = 2π√(m/k)

According to the law of conservation of momentum : (inelastic Collison)

m1 * u1 = (m1 + m2) * v

Where v is the final Velocity of the colliding bodies

0.008 * 280 = (0.008 + 0.992) * v

2.24 = 1 * v

v = 2.24m/s

K. E = P. E

K. E = 0.5mv^2

P.E = 0.5kx^2

0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2

0.5*1*5.0176 = 0.5*k*0.0225

2.5088 = 0.01125k

k = 2.5088 / 0.01125

k = 223.00444 N/m

Therefore,

Period (T) = 2π√(m/k)

T = 2π√(0.992+0.008) / 233.0444

T = 2π√0.0042910

T = 2π * 0.0655059

T = 0.4113772 s

6 0
3 years ago
A drop of oil of volume 10m it spread out on water to make a circular firm of radius 10m calculate the tickness of the firm
Effectus [21]

Answer:

h = 3.1 cm

Explanation:

Given that,

The volume of a oil drop, V = 10 m

Radius, r = 10 m

We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :

V=\pi r^2 h\\\\h=\dfrac{V}{\pi r^2}\\\\h=\dfrac{10}{\pi \times 10^2}\\\\h=0.031\ m\\\\h=3.1\ cm

So, the thickness of the film is equal to 3.1 cm.

8 0
2 years ago
A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul
Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

I = \frac{2}{5}mR^2

w= \frac{v}{R}

Now kinetic energy is given by

KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2

KE = \frac{7}{10}mR^2w^2

KE = \frac{7}{10}*150*(0.200)^2(50)^2

KE = 10500 J

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0
3 years ago
If the magnitude of a charge is half as much as another charge, but the force experienced is the same, then the electric field s
Kazeer [188]

Answer:

the electric field strength of this charge is two times the strength of the other charge

Explanation:

Using the relationship between electric field and the charge, which is inversely proportionality. Let the the magnitude of the first charge be Q and the respective electric field be E. It implies that;

E1/E2 = Q2/Q1

E2 = E1 x Q1/Q2

      = E x Q/ (Q/2)

       = 2E

8 0
2 years ago
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