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Inessa05 [86]
3 years ago
14

Difference between acceleration and decceleration​

Physics
1 answer:
mario62 [17]3 years ago
8 0

acceleration is considered to describe an increase or positive change of speed or velocity But deceleation is considered to describe a decrease or negative change of speed or velocity

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A) what is the light source for a plant growing in the shade?
Ulleksa [173]

a.) Plants that thrive in the shade are often able to hold on to sunlight for extensive periods of time; they're in a sense like the camels of the plaNt WoRld.

b.) Though artificial lights are not nearly as beneficial as the sun, one could invest in one of them plant growing light thingies, but sun-loving plants might be sad if u do this instead of letting them soak in ePic rays from the sun.

6 0
3 years ago
If the speed of a car is 20m/s .How long does it take to cover a distance of 1km?<br>​
3241004551 [841]

Answer: 50 seconds

Explanation:

1km=1000m

Distance/speed=time

1000/20=50

50 seconds

7 0
2 years ago
Read 2 more answers
Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P P and volume V V satisfy th
Verdich [7]

Answer:

the volume decreases at the rate of 500cm³ in 1 min

Explanation:

given

v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min

PV=C

vΔp + pΔv = 0

differentiate with respect to time

v(Δp/t) + p(Δv/t) = 0

(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0

40000 + 80kPa(Δv/t) = 0

Δv/t = -40000/80

= -500cm³/min

the volume decreases at the rate of 500cm³ in 1 min

3 0
3 years ago
What affects sounds at their source? MENTION DOPPLER EFFECT AND VIBRATIONS
wel

Answer:

Doppler effect, the apparent difference between the frequency at which sound or light waves leave a source and that at which they reach an observer, caused by relative motion of the observer and the wave source. This phenomenon is used in astronomical measurements, in Mössbauer effect studies, and in radar and modern navigation.

Explanation:

6 0
3 years ago
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To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

q(t) = e^{-\frac{t}{(R*C)}}

Here,

q = Charge

t = Time

R = Resistance

C = Capacitance

When the charge reach its half value it has passed 10ms, then the equation is,

\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}

Ln(\frac{1}{2}) = -\frac{0.01}{RC}

- RC = \frac{0.01}{Ln(1/2)}

RC = 0.014s

We know that RC is equal to the time constant, then

T = RC  = 0.014s = 14ms

Therefore the time constant for the process is about 14ms

5 0
3 years ago
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