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NARA [144]
3 years ago
13

Calculate 2 x 10^-3cm ÷ 2.5 x 10^4cm

Physics
1 answer:
Mamont248 [21]3 years ago
3 0
Here is your answer:

First find the notations:

2×10^-3
=0002
And...
2.5×10^4=25000

Then divide:

0002÷25000=8E-9

Your answer:
=8 x 10-8
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What is the speed of a wave that has a frequency of 173 Hz and a wavelength of 2.59 meters? Express your answer to the nearest w
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448 m/s is the correct answer.

Explanation:

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.A box falls to the ground from a delivery truck traveling at 30 m/s. After hitting the road, it slides 45 m to
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Answer:

t = 3 seconds

Explanation:

Given that,

Initial speed, u = 30 m/s

Final speed, v = 0

It slides 45 m to rest.it take the box to come to rest

We need to find how long it take the box to come to rest.

Let a be the acceleration and t is time.

v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(30)^2-u^2}{2(45)}\\\\=10\ m/s^2

Now finding time.

t=\dfrac{v-u}{a}\\\\t=\dfrac{30-0}{10}\\\\t=3\ s

So, the required time is 3 seconds.

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PLEASE HELP ME PLS PLS Imagine an object held at some height above the ground. It is released and falls toward the ground. Ignor
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Option e is true. The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

<h3>What is the law of conservation of energy?</h3>

According to the Law of conservation of energy. Although energy cannot be generated or destroyed, it may be transferred from one form to another.

The following statements are true;

a)All of its energy must be potential energy before it falls.

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2 years ago
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An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of
slega [8]

Answer:

T_{1}=94.9^{o}C

Explanation:

Given data

length=100mm

Diameter=5mm

Thermal conductivity=5 W/m.K

Power=50 W

Temperature=25°C

The temperature of heater surface follows from the rate equation written as:

T_{1}=T_{2}+\frac{q}{kS}

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

S=\frac{2\pi L}{ln(\frac{4L}{D} )} \\

Substitute the given values

S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m

The temperature of heater is then:

T_{1}=25^{o}C+\frac{50W}{5W/m.K*0.143m} \\T_{1}=94.9^{o}C

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

                           T_{1}=94.9^{o}C

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