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3 years ago

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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A cannon elevated at 40 degrees is fired at a wall 300m away on level ground. The initial speed of the cannonball is 89m/s How l

lianna [129]

For the answer to the question above, we'll have to use these formulas.

A) to find time to travel the 300m,
just find horizontal component of the velocity and divide.
ie x=89 x t x cos 40, t=x/89 x cos 40

<span>B) y=vtsin 40 - gt^2/2, just sub in
</span>
I believe you can do the rest.

I hope I helped you with my answers.

3 0

3 years ago

An ideal heat engine absorbs 97.2 kJ of heat and exhausts 83.8 kJ of heat in each cycle. What is the efficiency of the engine?

r-ruslan [8.4K]

Answer:

13.7%

Explanation:

Given that,

Heat absorbed by the engine = 97.2 kJ

Heat exhausted by the engine in each cycle = 83.8 kJ

We need to find the efficiency of the engine. It is calculated by the formula.

$\eta=1-\dfrac{Q_e}{Q_a}\\\\=1-\dfrac{83.8}{97.2}\\\\=0.137\\\\=13.7\%$

so, the efficiency of heat engine is 13.7%.

5 0

3 years ago

Neritic Sediments are deposited on the ocean floor in a sorted manor. In which order are the sediments ordered moving from the s

BaLLatris [955]

B <span> of Earth’s surface is covered by water. Very little or no light penetrates beyond a few hundred feet in water</span>

6 0

3 years ago

Tammy leaves the office, drives 34 km due north, then turns onto a second highway and continues in a direction of 35◦ north of e

Natasha2012 [34]

Answer:

The total displacement is 102 km $51^o$ north of east.

Explanation:

We can treat this problem as a trigonometric one, so we need to calculate the total displacement on the north and east.

$d_n=34km+79km*sin(35^o)=79km$

and

$d_e=79*cos(35^o)=65km$

The total displacement is given by:

$D=\sqrt{(79km)^2+(65km)^2}=102km$

with an agle of:

$\alpha =arctg(\frac{79km}{65km})=51^o$

4 0

3 years ago

Students are given the following information about a 2.0 kg, motorized toy boat on water. what net force is exerted on the boat

Natasha_Volkova [10]

The motorized toy boat experiences a net force of 0 N between 4 s and 8 s.

The motorized toy boat moves at 8 m/s (u) at 4 s and at 8 m/s (v) at 8 s. We can calculate the acceleration (a) in that period using the following kinematic expression.

$a = \frac{v-u}{\Delta t} = \frac{8m/s-8m/s}{8s-4s} = 0 m/s^{2}$

The object with a mass (m) of 2.0 kg experiences an acceleration of 0 m/s². We can calculate the net force (F) in that period using Newton's second law of motion.

$F = m \times a = 2.0 kg \times 0 m/s^{2} = 0 N$

The motorized toy boat experiences a net force of 0 N between 4 s and 8 s.

Learn more: brainly.com/question/13447525

6 0

3 years ago