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mojhsa [17]
2 years ago
7

A glass of root beer with a scoop of ice cream floating on top and a straw sticking out.

Physics
1 answer:
vampirchik [111]2 years ago
6 0

Answer:

These forces are all equal and cancel each other out. Gravity pushes downward on the ice cream. This can also be called the weight of the ice cream. Buoyant force pushes the ice cream upward

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Work=force*displacment
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stephen buys a new moped . he travels 4km south and then 6km east. how far does he need to go to get back where he started??
stiks02 [169]

Answer:

My answer is 7.2 km

Explanation:

When Stephen goes to the south and then to the east, he is drawing a right triangle, where the 4 km and 6 km sides are the cathetus of a right triangle.

Then we use the Pithagorean theorem to solve this problem. We need to find the hypotenuse.

c² = a² + b²

c² = 4² + 6²

c² = 16 + 36

c² = 52

c = 7.2 km

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3 years ago
At the moment t = 0, a 24.0-v battery is connected to a 5.00-mh coil and a 6.00-ω resistor. (a) immediately thereafter, how does
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3 years ago
Example 3 :
kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × \frac{0. 06}{0. 12* 1* 0. 9}

friction factor = 0. 52 × \frac{0. 06}{0. 108}

friction factor = 0. 52 × 0. 55

friction factor = 0. 289

b. When V = 3mls

Friction factor = 0. 52 × \frac{0. 06}{0. 12 * 3* 0. 9}

Friction factor = 0. 52 × \frac{0. 06}{0. 324}

Friction factor = 0. 52 × 0. 185

Friction factor = 0.096

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × \frac{1}{2*6. 6743 * 10^-11} × \frac{1^2}{0. 120} × \frac{1}{100}

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = 0. 096 × \frac{1}{1. 334 *10^-10} × \frac{3^2}{0. 120} × \frac{1}{100}

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

#SPJ1

4 0
1 year ago
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