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Alja [10]
1 year ago
10

Meme on newtons law of motion (should me made ur self not from any searh engine)

Physics
1 answer:
k0ka [10]1 year ago
8 0
He will be a pilot and he will fly the plane over bridges fewwww
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What statement best describes his motion as he jogs around the curved part of the track?
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Answer:

The answer would be A.

Explanation:

5 0
3 years ago
In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, s
serious [3.7K]

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

a = \dfrac{1.2 \times 10^5}{86}

a =1395.35\ m/s^2

Using equation of motion

v² = u² + 2 a s

s =\dfrac{v^2}{2a}

s =\dfrac{52^2}{2\times 1395.35}

s = 0.9689 m

The minimum depth of snow that would have stooped him is  s = 0.9689 m

8 0
3 years ago
Marco is looking for a used sports car,
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Answer:

B

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A

Explanation:

6 0
3 years ago
Amplitude modulation is used in _____.
spayn [35]

Amplitude modulation is a modulation technique used in electronic communication, most commonly for transmitting information via a radio carrier wave. In amplitude modulation, the signal strength of the carrier wave is varied in proportion to that of the message signal being transmitted. The message signal is, for example, a function of the sound to be reproduced by a loudspeaker, or the light intensity of pixels of a television screen. This technique contrasts with frequency modulation, in which the frequency of the carrier signal is varied, and phase modulation, in which its phase is varied.

AM was the earliest modulation method used to transmit voice by radio. It was developed during the first quarter of the 20th century beginning with Landell de Moura and Reginald Fessenden's radiotelephone experiments. It remains in use today in many forms of communication; for example it is used in portable two-way radios, VHF aircraft radio, citizens band radio, and in computer modems in the form of QAM. AM is often used to refer to mediumwave AM radio broadcasting.

6 0
3 years ago
Read 2 more answers
Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m/s^2, where t is in seconds. What
hjlf

To solve this problem we will use the given expression and derive it in order to find the algebraic expressions of velocity and position. These equations will be similar to those already known in the cinematic movement but will be subject to the previously given function. We start deriving the equation for velocity

a = 2t-6

\frac{dv}{dt} = 2t-6

Integrate acceleration equation

\int dv = (2t-6)dt

v = 2(\frac{t^2}{2})-6t+C_1

v=t^2-6t+C_1

At t = 0, v = 0

Replacing,

0 = 0^2-6*0+C_1

Therefore the value of the first Constant is

C_1 = 0

The expression can be escribed as,

v = t^2-6t

Calculate the velocity after 6s,

v=t^2-6t

v = 6^2-6*6

v = 0m/s

Now using the same expression we can derive the equation for distance

v = t^2-6t

\frac{dx}{dt} =t^2-6t

\int dx = \int (t^2-6t)dt

x = \frac{t^3}{3}-6\frac{t^2}{2}+C_2

At t=0, x=0

0 = \frac{0^3}{3}-6(\frac{0^2}{2})+C_2

Therefore the value of the second constant is

C_2 = 0

x = \frac{t^3}{3}-\frac{6t^2}{2}

Calculate the distance traveled after 11 s

At  t=11s

x = \frac{11^3}{3}-6(\frac{11^2}{2})

x = 80.667m

6 0
3 years ago
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