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Vesna [10]
3 years ago
12

Which of the following are NOT examples of digital information storage?

Physics
2 answers:
PolarNik [594]3 years ago
6 0
The answer would be D
VikaD [51]3 years ago
3 0
Vinyl records because it’s not a secure information storage
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Select the correct answer.
kherson [118]

Answer:

B chemical energy

Explanation:

4 0
3 years ago
what is the total distance traveled by a bike rider who rides for three hours at 40 km/hr and then two more hours at 50 km/hr?
marta [7]

Answer:

220 km

Explanation:

3 hours at 40 km/hr

40 km = 1 hour

40 km times 3 = 120 km

2 hours at 50 km/hr

50 km = 1 hour

50 times 2 = 100 km

Total distance

120 km + 100 km = 220 km

4 0
3 years ago
A 20-kg rock sits atop a 30 meter cliff. How much potential energy does it have?
Brut [27]
Gravitational potential energy ...

=  (mass) x (acceleration of gravity) x (height)

=  (20 kg) x (9.8 m/s²) x (30 m)

=   (20 x 9.8 x 30)  kg-m²/sec²

=   5,880 joules
6 0
3 years ago
The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid
Karolina [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

w=2250*2\pi *\frac{1}{60}\\ w=235.61

I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}

Now using Inertia an w

E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J

average power=\frac{11.4J}{0.230s}=50.2 W

b).

power=t*w

P=11.5465*0.25*235.61

P=2.957 kW

8 0
3 years ago
Read 2 more answers
A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a
Oliga [24]

Answer:

the mass drop by 6.5cm before coming to rest.

Explanation:

Given that:

the mass of the block M= 1.40 kg

angle of inclination θ = 30°

spring constant K = 40.0 N/m

mass of the suspended block m = 60.0 g = 0.06 kg

initial downward speed = 1.40 m/s

The objective is to determine how far does it drop before coming to rest?

Let assume it drops at y from a certain point in the vertical direction;

Then :

Workdone by gravity on the mass of the block is:

w_1g = 1.4*9.81*y*sin30 = - 6.867y

Workdone by gravity on the mass of the suspended block is:

w_2g = 0.06*9.81 = 0.5886

The workdone by the spring = -1/2ky²

= -0.5 × 40y²

= -20 y²

The net workdone is = -20 y² - 6.867y + 0.5886y

According to the work energy theorem

Net work done = Δ K.E

-20 y² - 6.867y + 0.5886 = 1/2 × 0.06 × 1.4²

-20 y²  - 6.867y + 0.5886 = 0.5 × 0.1176

-20 y²  - 6.867y + 0.5886 = 0.0588

-20 y²  - 6.867y + 0.5886 - 0.0588

-20 y²  - 6.867y + 0.5298 = 0

multiplying through by (-)

20 y²  + 6.867y - 0.5298 = 0

Using the quadratic formula:

\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}

where;

a = 20 ; b = 6.867 c= - 0.5298

\dfrac{-(6.867)  \pm \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}

= \dfrac{-(6.867)  + \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)} \ \ \ OR \ \ \  \dfrac{-(6.867)  - \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}= 0.0649 OR  −0.408

We go by the positive integer

y = 0.0649 m

y = 6.5 cm

Therefore; the mass drop by 6.5cm before coming to rest.

7 0
3 years ago
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