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Verizon [17]
3 years ago
11

Cole drives to school from home, starting from rest and accelerating for 10 minutes as he travels 6.0 km to school.

Physics
1 answer:
Firlakuza [10]3 years ago
3 0

Explanation:

this is the answer for your question. if you have any doubt.

you can send your doubt to:6369514784(what's app)

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A construction worker puts 20J of energy in to one strike of his hammer on the head of a nail. The energy transferred to driving
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efficiancy=40 percent

Explanation:

efficiency=energy output/energy input×100

efficiancy=8J/20J×100

efficiancy=0.4×100

efficiancy=40 percent

Mark brianliest if my answer suit your question..

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A trombone has a variable length. When a musician blows into the mouthpiece and causes air in the tube of the horn to vibrate, t
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Answer:

The frequency increases.

Explanation:

When the Musician draws the slide in the length of the horn gets shorter, which causes a decrease in the wavelength. A decrease in the wave length results in an increase in frequency.

Note:

The diameter of the horn has an effect on frequency, so a wider horn is effectively a long horn - open end correction ( distance between the the antinode and the open end of a pipe).

Frequency also depends on how hard the musician blows the trombone. The musician can change the frequency with the lip pressure being applied.

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3 years ago
A car accelerates uniformly from rest to a speed of 10m/s. Calculate its acceleration
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Give four examples of landforms where both the water and the land around it are flat.

Explanation:

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If you adjust the brightness of a light until it matches the intensity of sourness of the taste of a lemon, you are engaging in
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At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
lbvjy [14]

Answer:

F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

|F_T|=2\sqrt{34}k\frac{Q^2}{L}

\theta=tan^{-1}(\frac{5}{3})=59.03\°

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

and the magnitude is:

|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}

the direction is:

\theta=tan^{-1}(\frac{5}{3})=59.03\°

4 0
2 years ago
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