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3 years ago

11

It takes 100,832 J of work to lift an elevator 18.3 meters. If this is done in 21.0 seconds, what is the average power of the el

evator during the process? Explain your answer.

1 answer:

34kurt3 years ago

6 0

Answer:

4.80 kW

Explanation:

Power = work / time

P = 100,832 J / 21.0 s

P ≈ 4800 W

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Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th

aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0

3 years ago

Help me get a answer

daser333 [38]

Answer:

I think D sorry if I'm wrong

6 0

2 years ago

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$

$h = \frac{3}{4g}(v_1^2 -v_f^2)$

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

$h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m$

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0

3 years ago

A string that is 3.6 m long is tied between two posts and plucked. The string produces a wave that has a frequency of 320 Hz and

marin [14]

To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as

$\lambda = \frac{v}{f}$

Where,

v = Velocity

f = Frequency,

Our values are given as

L = 3.6m

v= 192m/s

f= 320Hz

Replacing we have that

$\lambda = \frac{192}{320}$

$\lambda = 0.6m$

The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,

$N = \frac{L}{\lambda}$

$N = \frac{3.6}{0.6}$

$N = 6$

Therefore the number of wavelengths of the wave fit on the string is 6.

5 0

3 years ago

When will an object dropped from rest attain a speed of 30 m/s?

stich3 [128]

<h3><u>Answer</u> :</h3>

Initial velocity = zero (i.e., free fall)

Final velocity = 30m/s

Acceleration due to gravity = 10m/s²

For a body falling freely under the action of gravity, g is taken positive $.$

◈ <u>First equation of kinenatics</u> :

⇒ v = u + gt

⇒ 30 = 0 + 10t

⇒ t = 30/10

⇒ <u>t = 3s</u>

Hence, object will attain a speed of 30m/s after 3s.

8 0

3 years ago