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mrs_skeptik [129]
3 years ago
8

When conducting an investigation outside, you can assume it is safe to touch my plant or animal you see.

Physics
1 answer:
noname [10]3 years ago
4 0

False, all scene are combed for clues and photographed.

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the maximum displacement of an oscillatory motion is A=0.49m. determine the position x at which the kinetic energy of the partic
kipiarov [429]

Answer:0.4 m

Explanation:

Given

Maximum displacement A=0.49

The sum of kinetic and elastic potential energy is \frac{1}{2}kA^2

where k=spring constant

U+K.E.=\frac{1}{2}kA^2

when K.E.=U/2

K.E.=kinetic energy

U=Elastic potential Energy

\rightarrow \ U+\frac{U}{2}=\frac{1}{2}KA^2\\\rightarrow \ \frac{3U}{2}=\frac{1}{2}KA^2\\\rightarrow \ U=\frac{1}{3}KA^2\\\rightarrow \ \frac{Kx^2}{2}=\frac{1}{3}KA^2\\\\x=\sqrt{\frac{2}{3}}A\\x=0.4\ m

3 0
3 years ago
A 4 kg and 6 kg bowling ball are dropped from the same height at the same time. The two balls strike the ground at the same time
Ray Of Light [21]
The force of gravity on objects is proportional to the mass of each object.

(That's a big part of the reason why, when you eat more and your mass
increases, you weigh more.)

The forces of gravity between the Earth and the 6kg ball are 50% greater
than the forces of gravity between the Earth and the 4kg ball.

(The gravitational forces between the 4kg ball and the 6kg ball, or between
both bowling balls and you, are so small that they may be ignored.)

5 0
3 years ago
Read 2 more answers
Two charges q1 and q2 exert a 90 N electrostatic force onto each other when they are 1 m apart. They are moved further away to a
Dovator [93]

Answer:

?????????????????????????

6 0
3 years ago
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The position of a 2.2 kg mass is given by x=(2t3â5t2) m, where t is in seconds.
Juliette [100K]
B b b b b b b bb  bb   bb b b b b b b b b b 
6 0
3 years ago
There's an electric field in some region of space that doesn't change with position. An electron starts moving with a speed of 2
tangare [24]

Answer:

Explanation:

Given

speed of Electron u=2\times 10^7\ m/s

final speed of Electron v=4\times 10^7\ m/s

distance traveled d=1.2\ cm

using equation of motion

v^2-u^2=2as

where v=Final velocity

u=initial velocity

a=acceleration

s=displacement

(4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}

a=5\times 10^{16}\ m/s^2

acceleration is given by a=\frac{qE}{m}

where q=charge of electron

m=mass of electron

E=electric Field strength

5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}

E=248.3\ kN/C                

5 0
3 years ago
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