All categories

4 years ago

50200 J of heat are removed from

Physics

1 answer:

Dmitry_Shevchenko [17]4 years ago

4 0

Correct Answer:

3.1375

Explanation:

Use equation $Q=mc$ Δ $T$ to find $m$

Plug in all variables $-50200=x\cdot 2000\cdot -8$

Answer: 3.1375

You might be interested in

Explain why fishes survive in ponds even when the temperature is below 0°c

zmey [24]

Answer:

After sufficient thickness of ice is formed it prevents further loss of heat from the bottom layers of water. This is why fishes and other aquatic animals and plants can survive in ponds and other water bodies even when the atmospheric temperature reaches or is well below 0 degrees.The anomalous expansion of water helps preserve aquatic life during very cold weather. When temperature falls, the top layer of water in a pond contracts becomes denser and sinks to the bottom. ... Thus, even though the upper layer are frozen, the water near the bottom is at 4°C and the fishes can survive in it easily.

3 0

3 years ago

What are the main characteristics of S.H.M

agasfer [191]

Answer:

•→ The motion of a particle or body in S.H.M acts towards a fixed point.

•→ Acceleration of the body under S.H.M is proportional to its displacement.

•→ This motion is periodic.

•→ Mechanical energy is conserved in S.H.M

Explanation:

S.H.M is Simple Harmonic Motion

$.$

4 0

3 years ago

Read 2 more answers

A book is thrown downward from the library window with a speed of 2.0\,\dfrac{\text m}{\text s}2.0 s m 2, point, 0, start frac

dem82 [27]

Answer: final Velocity v = 10.2m/s

Explanation:

Final speed v(t) is given as

v(t) = u + at .......1

Where; u = the initial speed

a = acceleration

t = time taken

The total distance travelled d is given as

d = ut + 1/2(at^2)

Given

d = 5.0m

u = 2.0m

a = g = 10m/s2 (acceleration due to gravity)

Substituting into the equation above we have

5 = 2t + 5t^2

5t^2 +2t -5 = 0

Applying the quadratic formula. We have;

t = 0.82s & t = -1.22s

t cannot be negative

t = 0.82s

From equation 1 above

v = 2.0m/s + 10(0.82)m/s

v = 10.2m/s

7 0

3 years ago

Read 2 more answers

A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial

Nataly_w [17]

Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

$qvB=m\frac{v^2}{r}$

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

$r=\frac{mv}{qB}$

For the ion of oxygen-16, we have:

$m_A=2.66\cdot 10^{-26}kg$

$q_A = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_A=2.90\cdot 10^6 m/s$

$B_A=1.30 T$

So the radius of its orbit is

$r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m$

For the ion of oxygen-18, we have:

$m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg$

$q_B = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_B=2.90\cdot 10^6 m/s$

$B_B=1.30 T$

So the radius of its orbit is

$r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m$

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

$d=2(r_B-r_A)=2(0.417-0.371)=0.092 m$

3 0

3 years ago

How the behavior of waves is affected by a medium

Schach [20]

The medium determines the speed of the wave traveling in it, which also can have a number of other effects, including how much the wave bends (refracts), whether it reflects, etc.
Because waves move through space, they must have a velocity. The velocity of a wave is a function of the type of wave, and the medium it travels through. Electromagnetic waves moving through a vacuum, for instance, travel at roughly 3 x
10
8
m/s. This value is so famous and common in physics it is given its own symbol, c.

3 0

3 years ago