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kirza4 [7]
3 years ago
9

A race car travels along a section of a track with a constant speed of 100 m/s. If this speed is maintained for 15 minutes, how

far does the car travel within that time? (a) 30 km (b) 60 km (c) 90 km (d) 120 km
Physics
1 answer:
liberstina [14]3 years ago
7 0
(C)90 km
100x60seconds x 15 minutes/1000metres
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According to the text, there is no energy shortage now, nor will there ever be. what reason (s) is given to support this stateme
Bas_tet [7]
The reason why there is no energy shortage nor will there ever be is because energy is being preserved and conserved and only changes form. It never gets lost or increased.
8 0
3 years ago
in positive numbers less than 1, the zeros between the decimal point and a non-zero number are _______ significant?
DedPeter [7]

Answer:

Explanation:

If a number of less than 1, then the number has a decimal point like

0.085, 0.008 e.t.c.

The zeros before the none zero digit are insignificant. The significant figure is 8 and 5.

But if there a zero between the none zero e.g. 0.0087056

Here the zero between 7 and 5 is significant, then the significant numbers are 8,7,0,5,6

But if the zero is not in between the none zero digit, then the zero is insignificant

E.g 0.05800

The last two zero is insignificant, the significant number is 5 and 8

So, If a positive numbers less than 1, the zeros between the decimal point and a non-zero number are NOT significant.

8 0
3 years ago
A communications channel has a bandwidth of 4,000 hz and a signal-to-noise ratio (snr of 30 db. what is the maximum possible dat
levacccp [35]
Through Shannon's Theorem, we can calculate the capacity of the communications channel using the value of its bandwidth and signal-to-noise ratio. The capacity, C, can be expressed as 

C = B × log₂(1 + S/N)

where B is the bandwidth of the channel and S/N is its signal-to-noise ratio.

Since the given SN ratio is in decibels, we must first express it as a ratio with no units as

SN (in decibels) = 10 × log (S/N)
30 = 10log(S/N)
log(S/N) = 3
S/N = 10³ = 1000

Now that we have S/N, we can solve for its capacity (in bits per second) as 

C = 4000 × log₂(1 + 1000) 
C = 39868.91 bps

Thus, the maximum capacity of the channel is 39868 bps or 40 kbps.

Answer: 40 kbps

 
4 0
3 years ago
It requires 49 J of work to stretch an ideal very light spring from a length of 1.4 m to a length of 2.9 m. What is the value of
nadya68 [22]

Answer:

44 N/m

Explanation:

The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m

The work needed to stretch a spring by <em>e</em> is given by

W = \frac{1}{2} ke^2

where <em>k</em> is spring constant.

k = \dfrac{2W}{e^2}

Using the appropriate values,

k = \dfrac{2\times 49\text{ J}}{1.5^2\text{ m}^2} = 43.55\ldots\text{ N/m} \approx 44\text{ N/m}

3 0
3 years ago
Read 2 more answers
A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan
mariarad [96]

Answer:

d = 9.69 cm

Explanation:

given,

mass of the block = 1.2 Kg

spring force constant(k) = 730 N/m

spring is compressed = d = ?

rough patch width = 5 cm

μ_k = 0.44

work done by friction = energy lost

\mu_k mg \times x = \dfrac{1}{2}kd^2-\dfrac{1}{2}mv^2

0.44\times 1.2 \times 9.8 \times 0.05 = \dfrac{1}{2}\times 730 \times d^2-\dfrac{1}{2}\times 1.2 \times 2.3^2

3.432 = 365 d^2

d = \dfrac{3.432}{365}

d = 0.0969 m

d = 9.69 cm

5 0
3 years ago
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