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2 years ago

The environment affects people, but people don't affect the environment.TRUE OR FALSE

Physics

1 answer:

Drupady [299]2 years ago

5 0

Answer: False

Explanation:

The statement that the environment affects people, but people don't affect the environment is incorrect. It should be noted that people also affect the environment.

Some of our impact on the environment include pollution, deforestation, overpopulation, burning of fossil fuels etc. Thus has resulted in changes such as erosion, climate change, etc.

You might be interested in

How far must 5N force pull a 50g toy car if 30J of energy are transferred?

Alborosie

Answer: 6 m

Explanation:

30 = 5 * d

d = 30/5

d = 6 m

7 0

1 year ago

An ice hockey puck is tied by a string to a stake in the ice. the puck is then swung in a circle. what force is producing the ce

Taya2010 [7]

In a circular motion scenario, the force that pulls the revolving object towards the centre is the force that produces the centripetal acceleration. So, in this case, the tension on the string is the force that pulls the puck towards the centre.

Therefore, it is the tension in the string that causes the centripetal acceleration of the puck

Hope I helped!! xx

8 0

2 years ago

Read 2 more answers

A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by

Airida [17]

Answer:

Mobility of the minority carriers, $\mu_{n} =1184.21 cm^{2} /V-sec$

Diffusion coefficient for minority carriers, $D_{n} = 29.20 cm^2 /s$

Verified from Einstein relation as $\frac{D_{n} }{\mu_{n} } = 25 mV$

Explanation:

Length of sample, $l_{s} = 2 cm$

Separation between the two probes, L = 1.8 cm

Drift time, $t_{d} = 0.608 ms$

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), $\mu_{n} = \frac{V_{d} }{E}$

Where the drift velocity, $V_{d} = \frac{L}{t_{d} }$

$V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s$

and the Electric field strength, $E = \frac{V}{l_{s} }$

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

$\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec$

The electron diffusion coefficient, $D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }$

$\triangle x = (\triangle t )V_{d}$ , where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

$\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm$

$D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s$

For the Einstein equation to be satisfied, $\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V$

$\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV$

Verified.

4 0

2 years ago

1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational

sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = $\frac{m v^{2} }{r}$

==> 1.1 × $10^{19}$ = (9.4 × $10^{21}$ × $v^{2}$ ) ÷ 1.5 × $10^{8}$

==> 1.1 × $10^{19}$ = $v^{2}$ × 6.26× $10^{13}$

==> $v^{2}$ = 1.1 × $10^{19}$ ÷ 6.26× $10^{13}$

==> $v^{2}$ = 0.17571885 × $10^{6}$

==> v= 0.419188323 × $10^{3}$ m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5× $10^{8}$ ÷ 0.419188323 × $10^{3}$

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0

2 years ago

Read 2 more answers

An automobile traveling at 25.0 km/h along a straight, level road accelerates to 65.0 km/h in 6.00 s. what is the magnitude of t

USPshnik [31]

Note that
1 km/h = (1000 m)/(3600 s) = 0.27778 m/s

Initial velocity, v₁ = 25 km/h = 6.9444 m/s
Final velocity, v₂ = 65 km/h = 18.0556 m/s

Time interval, dt = 6 s.

Calculate average acceleration.
a = (v₂ - v₁)/dt
= (18.0556 - 6.9444 m/s)/(6 s)
= 1.852 m/s²

Answer:
The average acceleration is 1.85 m/s² (nearest hundredth)

3 0

2 years ago