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yuradex [85]
3 years ago
14

Helphpphphphpphphphpphphpph

Physics
1 answer:
adell [148]3 years ago
7 0
Airplane with nose up: The plane's speed through the air is the square root of (80 m/s squared) plus (120 m/s squared. The whole picture is a right triangle, and the plane's speed is the hypotenuse. The angle is the angle whose tangent is (80/120). You can get it from a calculator, a book, a slide rule, or online from the site that rhymes with floogle. The man pulling the load is also a right triangle. The horizontal component is (hypotenuse) times (cosine of the angle). The vertical component is (hypotenuse) times (sine of the same angle). Fill in what you know, look up the sin and cos of 25 degrees and write those in too, and then you can solve for what you have to find.
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My mom and I walked 3,600 m in 90 minutes. What was our speed in m/min?
PtichkaEL [24]
Fill in the fraction: 3,600/90 = 40; turn it into a unit fraction.

40 mi/min
3 0
3 years ago
If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in
Sindrei [870]

Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm

5 0
3 years ago
An egg of mass 0.060 kg is dropped from the top of a building. Just before it reaches the ground, it has a total kinetic energy
3241004551 [841]

The velocity is 14 m/s

The parameters given on the question are

mass= 0.060 kg

kinetic energy= 5.9 joules

K.E= 1/2mv²

5.9= 1/2 × 0.060 × v²

5.9= 0.5 × 0.060v²

5.9= 003v²

v²= 5.9/0.03

v²= 196.66

v= √196.66

v= 14 m/s

Hence the velocity of the egg before it strikes the ground is 14 m/s

brainly.com/question/2084569?referrer=searchResults

3 0
2 years ago
If the force of attraction (gravity) on the moon is 1/6 that of the force on Earth, what would
Aloiza [94]

I believe that you would weigh around 68 or 69 N, or 7 kilograms.

4 0
2 years ago
If the mass is displaced 0.270 m from equilibrium and released, what is its speed when it is 0.130 m from equilibrium?
puteri [66]

Answer:

The speed is 1.52 m.

Explanation:

Given that,

Displacement =0.270 m

Distance = 0.130 m

Suppose a 0.321-kg mass is attached to a spring with a force constant of 13.3 N/m.

We need to calculate the angular velocity

Using formula of angular velocity

\omega=\sqrt{\dfrac{k}{m}}

Put the value into the formula

\omega=\sqrt{\dfrac{13.3}{0.321}}

\omega=6.43\ rad/s

We need to calculate the velocity

Using formula of velocity

v=\omega\sqrt{A^2-x^2}

Put the value into the formula

v=6.43\times\sqrt{0.270^2-0.130^2}

v=1.52\ m/s

Hence, The speed is 1.52 m.

6 0
3 years ago
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