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2 years ago

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(b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a c

onstant rate of 1 m/s, how fast is the area of the spill increasing when the radius is 26 m?

2 answers:

weeeeeb [17]2 years ago

8 0

Explanation:

Below is an attachment containing the solution.

Volgvan2 years ago

4 0

Answer:

163.4 or $52\pi$ m2/s

Explanation:

The rate of change of the radius is 1 m/s

$\dfrac{dr}{dt}=1$

The area of a circle is

$A=\pi r^2$

We differentiate this to get the rate of change of the area with the radius:

$\dfrac{dA}{dr}=2\pi r$

The rate of change of the area is

$\dfrac{dA}{dt} = \dfrac{dA}{dr}\times\dfrac{dr}{dt}=2\pi r \times1 = 2\pi r$

At r = 26 m,

$\dfrac{dA}{dt}=2\pi \times26=52\pi=163.4$

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A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the si

solmaris [256]

Answer:

4.384 * 10^13

Explanation:

Given the expression :

[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]

Applying the laws of indices

[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]

13.2733 * 10^19 ÷ 3.0276 * 10^6

(13.2733 / 3.0276) * 10^(19 - 6)

4.3840996 * 10^13

= 4.384 * 10^13

6 0

2 years ago

This is a question on my physics test :)

Licemer1 [7]

Answer:

119.6 J/Kg°C

Explanation:

Data obtained from the question include:

Mass of substance (ms) = 170 g

Initial temperature of substance (Ts) = 120 °C

Volume of water = 200 mL

Initial temperature of water (Ts) = 10 °C

Temperature of the mixture (T2) = 12.6 °C

Density of water = 1 g/mL

Specific heat capacity of water (Cw) = 4200J/Kg°C

Specific heat capacity of substance (Cs) =..?

Next, we shall determine the mass of water. This can be obtained as follow:

Volume of water = 200 mL

Density of water = 1 g/mL

Mass of water =..?

Density = mass /volume

1 = mass /200

Cross multiply

Mass of water = 1 x 200

Mass of water = 200 g

Convert 200 g of water to Kg

Mass of water = 200/1000 0.2 Kg

Mass of water = 0.2 Kg

Now, we obtained the specific heat capacity of the substance using the following formula:

MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0

Mass of water = 0.2 Kg

Initial temperature of water (Ts) = 10 °C

Specific heat capacity of water (Cw) = 4200J/Kg°C

Temperature of the mixture (T2) = 12.6 °C

Mass of substance (ms) = 170 g = 170/1000 = 0.17 Kg

Initial temperature of substance (Ts) = 120 °C

Specific heat capacity of substance (Cs) =..?

MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0

0.2× 4200(12.6 – 10) + 0.17×Cs×(12.6 – 120) = 0

840(2.6) + 0.17Cs(– 107.4) = 0

2184 – 18.258Cs = 0

Rearrange

2184 = 18.258Cs

Divide both side by the coefficient of Cs i.e 18258

Cs = 2184/18.258

Cs = 119.6 J/Kg°C

Therefore, the specific heat capacity of the substance is 119.6 J/Kg°C

7 0

2 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

<em>initial temperature of metals, = </em> $T_m$ <em />
<em>initial temperature of water, = </em> $T_i$ <em> </em>
<em>specific heat capacity of copper, </em> $C_p$ <em> = 0.385 J/g.K</em>
<em>specific heat capacity of aluminum, </em> $C_A$ = 0.9 J/g.K
<em>both metals have equal mass = m</em>

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

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6 0

2 years ago

What will the magnitude of the field be if the 10 nc charge is replaced by a 20 nc charge? Assume the system is big enough to co

MArishka [77]

Answer:

Same magnitude of the 10 nc charge cause the electric field is external.

Explanation:

To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.

As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.

F = qE

If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.

4 0

2 years ago

Two blocks A and B with mA = 2.6 kg and mB = 0.81 kg are connected by a string of negligible mass. They rest on a frictionless h

vladimir1956 [14]

Answer:

Explanation:

Given

Mass of block A $(m_a)$ =2.6 kg

Mass of block B $(m_b)$ =0.81 kg

Force of 6.5 N is applied on Block A

Force on Block A is F, Tension

thus

$F-T=m_a\times a$ -----1

where a=acceleration of the system

For Block B

$T=m_b\times a$ -----2

$F=m_b\times a+m_a\times a$

$F=\left ( m_a+m_b\right )a$

$F=\left ( 2.6+0.81\right )a$

$a=\frac{6.5}{3.41}$

$a=1.96 m/s^2$

$T=0.81\times 1.96=1.54 N$

5 0

3 years ago