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3 years ago

5

What is the potential energy of a 45 kg object resting on the ground?

1 answer:

yKpoI14uk [10]3 years ago

3 0

Answer:The potential energy is zero

Explanation:

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A cube of plastic 1.2x10^-5 km on a side has a mas of 1.1 g. what is its density in g/cm^3?

Mamont248 [21]

Density = mass / volume

mass = 1.1 g

volume = length of side ^ 3 = [1.2 * 10^-5 km * 100000 cm/km]^3 = [1.2 cm]^3 = 1.728 cm^3

density = 1.1 g / 1.728 cm^3 = 0.64 g / cm^3

5 0

3 years ago

an elevator mass of 7700 kg falls from a height of 32 m after a sudden failure in the hoisting cable. The mass is stopped by a s

valkas [14]

Answer:k=28.29 kN/m

Explanation:

Given

mass $m =7700 kg$

height from which Elevator falls $h=32 m$

Let x be the compression in the spring

thus From conservation of Energy Potential energy will convert in to Elastic Potential Energy of spring

$\frac{kx^2}{2}=mg(h+x)$ ----------1

also maximum acceleration is 5g

thus

$mg-kx=ma$

here $a=-5g$

$kx=mg-m(-5g)=6mg$

$x=\frac{6mg}{k}$

Substitute x in equation 1

$0.5\times k\times (\frac{6mg}{k})^2=mg(h+\frac{6mg}{k})$

$18\frac{(mg)^2}{k}=mgh+6\frac{(mg)^2}{k}$

$k=12\cdot \frac{mg}{h}$

$k=12\times \frac{7700\times 9.8}{32}$

$k=28.29 kN/m$

4 0

3 years ago

A cathode ray tube is made of glass with a small amount of some kind of gas in it. It has metal electrodes at each end to pick u

LuckyWell [14K]

Scientists could investigate this idea by making cathode ray tubes out of different materials to see if the ray was the same.

6 0

2 years ago

Read 2 more answers

Ok plz help this is my entrance exam question for physics in grade 9 (yes i am moving to a new school again)

vagabundo [1.1K]

Answer:

300 W

Explanation:

We know that

P = W/t

P = mgh/t

P = 40 × 10 × 3/4

P = 10 × 10 × 3

P = 300 W

4 0

3 years ago

The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the

Ivanshal [37]

Answer:

(a) $\alpha = - 1.32\ rev/m^{2}$

(b) $\theta = 13674\ rev$

(c) $\alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}$

(d) $a = 22.458\ m/s^{2}$

Solution:

As per the question:

Angular velocity, $\omega = 190\ rev/min$

Time taken by the wheel to stop, t = 2.4 h = $2.4\times 60 = 144\ min$

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

$\omega' = \omega + \alpha t$

$omega'$ = final angular velocity

$\omega$ = initial angular velocity

$\alpha$ = angular acceleration

Now,

$0 = 190 + \alpha \times 144$

$\alpha = - 1.32\ rev/m^{2}$

Now,

(b) The no. of revolutions is given by:

$\omega'^{2} = \omega^{2} + 2\alpha \theta$

$0 = 190^{2} + 2\times (- 1.32) \theta$

$\theta = 13674\ rev$

(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

$\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}$

(d) The radial acceleration is given by:

$\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s$

Linear acceleration is given by:

$a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}$

$a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}$

5 0

2 years ago