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2 years ago

Calculate the speed for a car that went a distance of 125 miles in

1 answer:

3 0

Speed = distance / time
= 125 miles / 2 hr
= 62.5 mi/h

Convert 65.5 mi/h to km/h

= 65.5 / 5
= 13.1
=13.1 x 8
= 104.8 km/h

Therefore speed equal 104.8 km/h

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Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B

WARRIOR [948]

$\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}$

Explanation:

Given:

$\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}$

$\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}$

The cross product $\textbf{A}×\textbf{B}$ is given by

$\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|$

$= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}$

$= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}$

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3 years ago

In figures (a)-(d), a block moves to the right in the positive x direction through the displacement Δx while under the influence

Elena L [17]

-------ca>a>b>s---c>a>d>b©©s>c>a>b

5 0

3 years ago

Heat transfer that doesn’t need a medium?<br><br> A) Radiation<br> B)Convection<br> C)Conduction

Degger [83]

Radiation is a method of heat transfer through electromagnetic waves. This transfer does not require a medium in order for heat to be transferred.

The solution is A.

8 0

3 years ago

Read 2 more answers

Equation of uniformly accelerated motion

Romashka [77]

This means acceleration a is constant.

Let

a) vo be the initial speed, at t=0

b) v be the final speed after time t

c) d distance travelled in time t

Then we have:

a) v=vo+a×t

b) v²=vo²+2×a×d (Galilei's equation)

c) d=vo×t+a×t²/2

d) average speed vm=(vo+v)/2

3 0

3 years ago

Question: Self-test 3.12 Calculate the change in G for ice at -10°C, with density 917 kg mº, when the pressure is increased from

Akimi4 [234]

The change in the Gibb's free energy per mole (G) is 1.96 J.

The given parameters:

Density of the ice, ρ = 917 kg/m³
Initial pressure, P₁ = 1.0 bar
Final pressure, P₂ = 2.0 bar
Temperature, T = - 10 C
Mass of water = 18 g

The change in the Gibb's free energy per mole (G) is calculated as follows;

$\Delta G = V(P_2-P_1) \\\\$

where;

V is the volume of the ice

$Density = \frac{Mass}{Volume} \\\\Volume = \frac{Mass}{Density} \\\\Volume = \frac{18 \times 10^{-3} \ kg}{917 \ m^3} \\\\Volume = 1.96 \times 10^{-5} \ m^3\\\\Volume = 1.96 \times 10^{-5} \ m^3 \times \frac{1000 \ L}{m^3} \\\\Volume = 0.0196 \ L$

Change in pressure;

$P_2 - P_1 = 2.0 \ bar \ - \ 1.0 \ bar = 1.0 \ bar = 0.987 \ atm$

The change in the Gibb's free energy per mole (G);

$\Delta G= V(P_2-P_1)\\\\\Delta G = 0.0196\ L \times 0.987\ atm \\\\\Delta G = 0.0193 \ L.atm\\\\1 \ L.atm = 101.325 \ J\\\\\Delta G = 0.0193 \ L.atm \times \frac{101.325 \ J}{1 \ L.atm} \\\\\Delta G = 1.96 \ J$

Thus, the change in the Gibb's free energy per mole (G) is 1.96 J.

Learn more about Gibb's free energy here: brainly.com/question/10012881

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3 years ago