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3 years ago

14

This demonstrates which of the following?

2 answers:

LekaFEV [45]3 years ago

5 0

Answer:

A. is the answer

Papessa [141]3 years ago

3 0

I think the answer is 8!

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A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i

galben [10]

Answer:

350 F to 100 F it take approx 87.33 min

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling

$\frac{dT}{dt}$ = -k(T-Ta)

$\frac{dy}{dt}$ = $\frac{d}{dt}$ (T(t) -Ta)

= $\frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt}$ = -k(T-Ta)

-ky $\frac{dy}{dt}$ = -ky

T(t) -Ta = (To -Ta) $e^{-kt}$ T(t) = Ta+ (To -Ta) $e^{-kt}$

put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) $e^{-k30}$

k = 0.025575

so here for T(t) = 100F

100 = 70 + ( 350 - 70 ) $e^{-0.025575*t}$

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min

5 0

4 years ago

A 50-V de voltage source was connected in series with a resistor and capacitor. Calculate the current in A to two significant fi

Naddik [55]

Answer:

Current through circuit will be $0.2706\times 10^6A$

Explanation:

We have given source voltage v = 50 volt

Resistance $R=25Mohm=25\times 10^6ohm$

Capacitance $C=0.1\mu F=0.1\times 10^{-6}F$

Time t = 5 sec

Time constant of RC circuit is given by $\tau =RC=25\times 10^6\times 0.1\times 10^{-6}=2.5sec$

We know that voltage across capacitor is given by $v_c=v_s(1-e^{\frac{-t}{\tau }})$

$v_c=50(1-e^{\frac{-5}{2.5 }})=43.2332v$

So current will be $=\frac{v_s-v_c}{R}=\frac{50-43.2332}{25\times 10^{-6}}=0.2706\times 10^6A$

So current through circuit will be $0.2706\times 10^6A$

4 0

4 years ago

A car accelerates uniformly from rest to 20 m/sec in 5.6 sec along a level stretch of road. Ignoring friction, determine the ave

bonufazy [111]

Answer:

(a) $P=33000W$

(b) $P=51000W$

Explanation:

The average power is defined as the amount of work done during a time interval:

$P=\frac{W}{t}(1)$

According to work-energy theorem, the work done is equal to the change in kinetic energy. So, we have:

$W=\Delta K\\W=K_f-K_0\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\(2)$

Recall that the weight is given by:

$w=mg\\m=\frac{w}{g}(3)$

The car accelerates uniformly from rest ( $v_0=0$ ). Replacing (3) in (2), we have:

$W=\frac{wv_f^2}{2g}$

(a) Finally, we replace this in (1):

$P=\frac{wv_f^2}{2gt}\\P=\frac{9000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=33000W$

(b)

$P=\frac{14000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=51000W$

3 0

4 years ago

What does the slope of a secant line represent

blsea [12.9K]

Answer:

Secant line is a straight line which joins the two points of functions and it also represents the slope between two points.

Explanation:

Secant line is also known as average rate of change of functions between two points and also called slope.

In mathematical form,

Secant line = average rate of change = slope

Slope m = y/x

3 0

3 years ago

A 14.0 g wad of sticky clay is hurled horizontally at a 110 g wooden block initially at rest on a horizontal surface. The clay s

ahrayia [7]

Answer:

86.53 m/s

Explanation:

Given:

Mass of clay (m) = 14.0 g = 0.014 kg

Mass of block (M) = 110 g = 0.110 kg

Initial speed of block (U) = 0 m/s

Sliding distance (d) = 7.50 m

Coefficient of friction between block and surface (μ) = 0.650

Let the initial speed of clay be 'u' and speed of clay and block just after collision be 'v'.

Now, momentum is conserved just before and just after collision.

Momentum just before collision = $mu + 0 = mu$

Momentum just after collision = $(m + M)v$

Therefore, $mu=(M+m)v$ --------- (1)

Now, using newton's second law and we find the acceleration of the system.

The frictional force is given as:

$f=\mu mg=-ma\\\\a=-\mu g$

Now, using equation of motion, we can find the velocity just after collision.

$0^2=v^2+2ad\\\\v=\sqrt{-2ad}\\\\v=\sqrt{-2\times (-\mu g)\times d}\\\\v=\sqrt{2\mu gd}$

Plug in the given values and find 'v'. This gives,

$v=\sqrt{2\times 0.650\times 9.8\times 7.50}\\\\v=\sqrt{95.55}=9.77\ m/s$

Now, using equation (1) and substituting the given values, we get:

$0.014u=(0.014+0.110)\times 9.77\\\\u=\frac{0.124\times 9.77}{0.014}\\\\u=86.53\ m/s$

Therefore, the speed of the clay immediately before impact is 86.53 m/s.

7 0

4 years ago