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3 years ago

PHYSICSwhat is the cos? given the triangle above? please show work

Physics

1 answer:

Arlecino [84]3 years ago

8 0

The cosine of the angle in a right triangle is (adjacent side) divided by (hypotenuse).

As luck would have it, the picture doesn't give us the length of the adjacent side, so we have to calculate THAT before we can calculate the cosine.

You can easily find the length of the adjacent side ... call Pythagoras in. The adjacent side is the square root of (10-squared minus 2-squared).

When you have that, the cosine of the angle is just (that) divided by 10.

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A heavy dart and a light dart are launched vertically by identical ideal springs. Both springs were initially compressed by the

sveticcg [70]

Answer:

Explanation:

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2 years ago

We normally write our address as a street number in a city or town, which is in a state or province, followed by a country. How

enot [183]

Answer:

Earth, Solar system, Milky way galaxy,Local group, super cluster, Universe.

Explanation:

We usually write our address starting from house no then street no then the town or city and after that comes the province and country. So we write address starting from smaller to larger. Let's do it in this case.First comes the planet earth which lies in the solar system. So our solar system is a smaller part of the milky way galaxy. Milky way galaxy is just a one galaxy in the cluster of galaxies which is called the local group.Local group is further a part of a super cluster. Then comes our universe which consists of 10 million super clusters.

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2 years ago

A cylinder contains 3.0 L of oxygen at 310 K and 2.5 atm. The gas is heated, causing a piston in the cylinder to move outward. T

Alex_Xolod [135]

Answer:

The gas pressure is: 1.55 atm.

Explanation:

We need to use the equation that relate the variables given at the exercise (pressure, temperature and volume) from the ideal gas law formula, when the mass is constant we can reduce the expretion $PV=nRT$ to $\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2} }$ solving to P2 we get: $\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}=P_{2}$ replace the values $P_{2}=\frac{2.5*3*610}{9.5*310} =1.55(atm)$ .

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2 years ago

A 5.0 kg cannonball is fired horizontally at 68 m/s from a 15-m-high cliff. A strong tailwind exerts a constant 12 N horizontal

jeka57 [31]

Answer:

3.7 m

Explanation:

ASSUMING this means extra distance beyond where the cannonball would land WITHOUT the wind assistance but in general ignoring air resistance. Hmmmmmm...tricky

The ball drops from vertical rest to ASSUMED horizontal ground 15 m below in a time of

t = √ (2h/g) = √(2(15)/9.8) = 1.75 s

Without the tail wind, the ball travels horizontally

d = vt = 68(1.75) = 119 m

The tailwind exerts a constant acceleration on the ball of

a = F/m = 12/5.0 = 2.4 m/s²

The average horizontal velocity during the flight is

v(avg) = (68 + (68 + 2.4(1.75)) / 2 = 70.1 m/s

so the distance with tailwind is

d = v(avg)t = 70.1(1.75) = 122.675 m

The extra distance is 122.675 - 119 = 3.675 = 3.7 m

8 0