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WINSTONCH [101]
4 years ago
7

a jet fighter accelerates at 17.7 m/s^2 , increasing its velocity from 119 m/s to 233 m/s. how much time does that take?

Physics
1 answer:
Aleksandr-060686 [28]4 years ago
7 0

Answer:

6.44 s

Explanation:

Given:

v₀ = 119 m/s

v = 233 m/s

a = 17.7 m/s²

Find: t

v = at + v₀

(233 m/s) = (17.7 m/s²) t + (119 m/s)

t = 6.44 s

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An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s
gizmo_the_mogwai [7]

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

Q = ms\Delta T + mL

here we have

Q = 40(950)(680 - 32) + 40(450 \times 10^3)

Q = 24624 kJ + 18000 kJ

Q = 42624 kJ

Since this is the amount of aluminium per hour

so power required to melt is given by

P = \frac{Q}{t}

P = \frac{42624}{3600} kW

P = 11.84 kW

Since the efficiency is 85% so actual power required will be

P = \frac{11.84}{0.85} = 13.93 kW

Part B)

Total energy consumed by the furnace for 30 hours

Energy = power \times time

Energy = 13.93 kW\times 30 h

Energy = 417.9 kWh

now the total cost of energy consumption is given as

R = P \times 20 \frac{Cents}{kWh}

R = 417.9 kWh\times  20 \frac{cents}{kWh}

R = 8357.6 Cents

3 0
3 years ago
A bike rider pedals with constant acceleration to reach a velocity of 7.5 (vf)m/s over a time of 4.5 s(t). during the period of
BigorU [14]
<span>The initial velocity of the bike was 1.67 (vf)m/s. This is found by evaluating 7.5/4.5 which yields the velocity per unit of time which is equivalent to initial velocity.</span>
6 0
4 years ago
When a rubberband is stretched all the way back, it is an example of which type of energy?
avanturin [10]
The answer is C. elastic potential energy
5 0
3 years ago
Read 2 more answers
If the torque required to loosen a nut that
liraira [26]

Explanation:

τ = Fr

34 Nm = F (0.30 m)

F ≈ 113 N

7 0
3 years ago
Question 8
viktelen [127]

Answer: D(t) = 8.e^{-0.4t}.cos(\frac{\pi }{6}.t )

Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:

y = a.sin(\omega.t) or y = a.cos(\omega.t)

where:

|a| is initil displacement

\frac{2.\pi}{\omega} is period

For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:

y=a.e^{-ct}.cos(\omega.t) or y=a.e^{-ct}.sin(\omega.t)

For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:

y=a.e^{-ct}.cos(\omega.t)

period = \frac{2.\pi}{\omega}

12 = \frac{2.\pi}{\omega}

ω = \frac{\pi}{6}

Replacing values:

D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)

The equation of displacement, D(t), of a spring with damping factor is D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t).

3 0
3 years ago
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