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Burka [1]
3 years ago
11

Pls help pls due today

Physics
2 answers:
BaLLatris [955]3 years ago
7 0
The mutualism I believe. So sorry if I’m wrong
Zina [86]3 years ago
5 0
Try going to the video under the question see if it answers it in there if not so sorry
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NEED HELP ASAP!!!!
podryga [215]
The answer is B
I rhink
6 0
3 years ago
An object with a momentum of 1500 kg-m/s directed east is acted upon by an impulse of 100.0 kg-m/s in the same direction. What i
fiasKO [112]

Answer:

The final momentum of the body = 1600 kgm/s

Thus the impulse that acted on the body. = 500 N.s

Explanation:

Momentum: This can be defined as the product of mass and velocity. The S.I unit of momentum is kgm/s. Momentum is a vector quantity.

Mathematically, momentum can be expressed as

M =mv.

Where M = momentum of the object, m = mass of the object, v = velocity of the object.

Impulse acting on the object = Final momentum of the object - initial momentum of the object

I = M₂ - M₁

M₂ = I + M₁......................... Equation 1

Where I = impulse, M₁ and M₂ = Final and initial momentum.

Note:

(i) The momentum and impulse act in the same direction

(ii) impulse is also a vector quantity.

Given: M₁= 1500 kgm/s, I = 100 kgm/s.

Substituting these values into equation 1

M₂ = 1500 + 100

M₂ = 1600 kgm/s.

Thus the final momentum of the body = 1600 kgm/s

14.

I = m(v-u)............................................... Equation 2.

Where I = impulse on the object, m = mass of the object, v = final velocity, u = initial velocity.

Given: m = 100 kg, v = 15 m/s, u = 10 m/s.

I = 100(15-10)

I = 100(5)

I = 500 N.s or 500 kgm/s.

Thus the impulse that acted on the body. = 500 N.s

7 0
4 years ago
At a certain instant, a rotating turbine wheel of radius RR has angular speed ωω (measured in rad/srad/s). What must be the magn
jonny [76]

Answer:

Explanation:

The magnitude of the acceleration makes an angle of 30° with the tangential velocity.

Resolving the acceleration to tangential and radial acceleration

at = aCos30 = √3a/2

ar = aSin30 = ½a

a = 2•ar

Then, the tangential acceleration is the linear acceleration, so the relationship between the tangential acceleration and angular acceleration is given as:

at = Rα

Then, α = at/R

since at = √3a/2

Then, α = √3 at/2R, equation 1

The radial acceleration is given as

ar = ω²R

Note that, at² + ar² = a²

at = √(a²-ar²)

Back to equation 1

α = √3 at/2R

α = √3√(a²-ar²)/2R

α = √3√(a²-(w²R)²)/2R

α = √3(a²-w⁴R²) / 2R

Also, a = 2•ar = 2w²R

Then,

α = √3((2w²R)²-w⁴R²) / 2R

α = √3(4w⁴R²-w⁴R²) / 2R

α = √3(3w⁴R²) / 2R

α = √9w⁴R² / 2R

α = 3w²R / 2R

α = 3w²/2

7 0
3 years ago
a boat sails along the shore. to an observer, the boat appears to move at a speed of 3m/s, and a man on the boat walking foward
lina2011 [118]
It is 6.7s. you need to slove the promble
6 0
3 years ago
Why do boys like boys
Llana [10]

Answer:

maybe they're gay maybe they act gay so they can be around there girl crush cause a lot of girls like to hang out around gay guys

Explanation:

4 0
3 years ago
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