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1 year ago

How much heat is released when 432 g of water cools down from 71'c to 18'c?

1 answer:

7 0

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

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A biophysics experiment uses a very sensitive magnetic field probe to determine the current associated with a nerve impulse trav

fenix001 [56]

Answer:

The peak current carried by the axon is 5.85 x 10⁻⁸ A

Explanation:

Given;

distance of the field from the axon, r = 1.3 mm

peak magnetic field strength, B = 9 x 10⁻¹² T

To determine the peak current carried by the axon, apply the following equation;

$B = \frac{\mu I}{2\pi r}$

where;

B is the peak magnetic field

r is the distance of the magnetic field from axon

μ is permeability of free space = 4π x 10⁻⁷

I is the peak current

Re-arrange the equation and solve for "I"

$B = \frac{\mu I}{2\pi r} \\\\I = \frac{B*2\pi r}{\mu} \\\\I = \frac{9*10^{-12}*2*\pi *1.3*10^{-3}}{4\pi *10^{-7}} \\\\I = 5.85 *10^{-8} \ A$

Therefore, the peak current carried by the axon is 5.85 x 10⁻⁸ A

7 0

1 year ago

An ice rescue team pulls a stranded hiker off a frozen lake by throwing him a rope and pulling him horizontally across the essen

Anuta_ua [19.1K]

Answer:T=116.84 N

Explanation:

Given

Weight of hiker =1040 N

acceleration $a=1.1 m/s^2$

Force exerted by Rope is equal to Tension in the rope

$F_{net}=T=ma_{net}$

$T=\frac{1040}{g}\times 1.1$

$T=116.84 N$

8 0

2 years ago

Military specifications often call for electronic devices to be able to withstand accelerations of 10 g. to make sure that their

trapecia [35]

The solution for this problem is:
(10 x 9.8) = 98.1 m/sec^2 acceleration. Time, to travel 9.4cm or (.094m.), at acceleration of 98m/sec^2

= sqrt(2d/a), = sqrt (98.1 m/sec^2/0.094m) = 32.3050619 sec per cycle

Frequency = (w/2pi), = 32.3050619/2pi
= 32.3050619/6.28318531
= 5.14 Hz would be the answer

6 0

1 year ago

A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal

Nataly_w [17]

Answer:

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature $T_i$ = 93.2°C

We know

$Q = m c \Delta T$ ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in $J/g^{\circ}C$

$\Delta T$ = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal = $125\times 0.9\times (93.2-T_f)$

Heat gained by the water = $100\times 4.184\times(T_f -18.3)$

thus, we have

$125\times 0.9\times (93.2-T_f)$ = $100\times 4.184\times(T_f -18.3)$

$10485-112.5T_f = 418.4T_f - 7656.72$

⇒ $T_f = 34.17^oC$

Therefore, the final temperature will be = 34.17°C

6 0

2 years ago

What is the magnitude of the relative angle φ

melomori [17]

Complete question is;

A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².

What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦

Answer:

14.08°

Explanation:

The time covered will be given by the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Now, the slope of the flight path at the point of impact will be given by the formula;

tan α = V_y/V_x

We are given V_x = 24 m/s

V_y will be gotten from the formula;

v = gt

Thus;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Thus ;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°

6 0

1 year ago