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Lunna [17]
2 years ago
10

What is the momentum of a bird with a mass of 0.018kg flying at 15m/s?

Physics
1 answer:
Mazyrski [523]2 years ago
6 0

Answer:

0.27 kg-m/s

Explanation:

i believe this is the correct answer

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Calculate (a) the torque, (b) the energy, and (c) the average power required to accelerate Earth in 4.0 days from rest to its pr
natima [27]
<h2>Answer:</h2>

Torque = <em>2.05 x 10²⁸ Nm</em>

Energy = <em>3.54 x 10³³ J</em>

Average power = <em>1.02 x 10²⁸ W</em>

<h2>Explanation:</h2>

(a) Torque (τ) is the rotational effect of a given force.  

It is given by

τ = I x α          -------------(i)

Where;

I = rotational inertia of the object

α = angular acceleration of the object.

In this case, the object is the Earth. Therefore,

I = 9.71 x 10³⁷ kg m²

α = ω / t

Where;

ω = angular velocity of earth = 2π rad / day

<em>Since </em>

<em>1 day = 24 hours and 1 hour = 3600seconds</em>

<em>1 day = 24 x 3600 seconds = 86400seconds</em>

<em>=> ω = 2π rad / 86400seconds</em>

<em>=> ω = 7.29 × 10⁻⁵ rad/s</em>

<em />

t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds

=> α = ω / t

=> α = 7.29 × 10⁻⁵ / 345600

=> α = (7.29 × 10⁻⁵) / (3.456 x 10⁵)

=> α = (7.29 × 10⁻⁵⁻⁵) / (3.456)

=> α = (7.29 × 10⁻¹⁰) / (3.456)

=> α = 2.11 × 10⁻¹⁰ rad/s²

Now substitute the values of I and α into equation (i)

τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰

τ = 9.71 x 10²⁷ x 2.11

τ = 20.5 x 10²⁷ Nm

τ = 2.05 x 10²⁸ Nm

(ii) The energy (rotational energy) E is given by;

E = \frac{1}{2} x I x ω

E = \frac{1}{2} x 9.71 x 10³⁷ x 7.29 × 10⁻⁵

E = 35.4 x 10³² J

E = 3.54 x 10³³ J

(iii) The average power P, is given by;

P = E / t

P = 3.54 x 10³³ / 345600

P = 1.02 x 10²⁸ W

5 0
3 years ago
The table below describes some features of methods used to generate electricity. Name method 4.
alina1380 [7]

Answer:

Hydroelectricity

Explanation:

Because of flooding of water, we can assume that the electricity was generated by Water which is known as Hydroelectricity.

4 0
2 years ago
A 94 g particle undergoes SHM with an amplitude of 8.3 mm, a maximum acceleration of magnitude 7.8 x 103 m/s2, and an unknown ph
Lelechka [254]

Answer:

a) T = 6.49*10^-3 s

b) v = 8 m/s

c) E = 3 J

d) F = 733 N

e) F = 366.5 J

Explanation:

Given

Mass of particle, m = 94 g = 0.094 kg

Amplitude of the particle, A = 8.3 mm = 8.3*10^-3 m

Maximum acceleration of particle, a = 7.8*10^3 m/s²

the equation describing Simple Harmonic Motion is given as

x = A cos (wt +φ)

To fond the acceleration of this relationship, we would have to integrate. Twice, the first would be a Velocity, and the second acceleration that we need.

Velocity = dx/dt = -Aw sin(wt + φ)

Acceleration = d²x/dt = -Aw² cos(wt + φ)

From the question, we were given, magnitude of acceleration to be 7.8*10^3 m/s²

Aw² = 7.8*10^3

w² = 7.8*10^3 / A

w² = 7.8*10^3 / 8.3*10^-3

w² = 939759

w = √939759

w = 969

Recall, T = 2π/w, so that

T = (2 * 3.142) / 969

T = 6.49*10^-3 s

Maximum speed = Aw

Maximum speed = 8.3*10^-3 * 969

Maximum speed = 8.0 m/s

Total mechanical energy oscillator =

mgx + 1/2mx² =

1/2mv(max)² =

1/2 * 0.094 * 8² =

3 J

Maximum displacement

x = A cos(wt + φ)

For x to be maximum here, then cos(wt + φ) Must be equal to 1

Acceleration = d²x/dt² = -Aw²

And force = mass * acceleration

Force = 0.094 * 7.8*10^3

Force = 733 N

x = A cos(wt + φ), where cos(wt + φ) = 1/2

d²x/dt² = -Aw² * 1/2

d²x/dt² = 733 * 0.5

= 366.5 N

7 0
3 years ago
Can work done=mass*acceleration*displacement(work=m*a*s)
Airida [17]

no, work is = force * distance or displacement


5 0
3 years ago
If the wave speed stays the same, which of the following decreases as the frequency increases?
dalvyx [7]

Answer:yes

Explanation:

8 0
3 years ago
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