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2 years ago

Which of the following best characterizes the field of physics

Physics

1 answer:

Sav [38]2 years ago

7 0

Would love to help you but there are no options for me to choose from

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You raise a bucket of water from the bottom of a deep well. if your power output is 138 w , and the mass of the bucket and the w

inn [45]

Maybe 34.5

I'm not quite sure

8 0

3 years ago

Two electric charges are moved so that they are twice as far apart as they had originally been. Is the force they experience fro

Natasha_Volkova [10]

Answer:

No.

Explanation:

The force that two particle experience is inversely proportional to the sqare of the distance, this is:

$F \ \alpha \ \frac{1}{D^{2}}$ for a distance D

If we move them so that D is doubled:

$\frac{1}{2^{2}.D^{2} }$ = $\frac{1}{4} \eq \frac{1}{.D^{2} } \eq$

Then the force they experience is one fourth of the original.

5 0

2 years ago

An ideal gas occupies 600 cm3 at 20c. at what temperature will it occupy 1200 cm3 if the pressure remains constant? 10c 40c 100c

Anna11 [10]

ANS : 313℃
You need to use K in this.
To convert ℃ to Kelvin (K), add 273.15 to ℃.

5 0

3 years ago

A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second

Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

$v_x = 15 m/s$

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

$d= v_x t =(15 m/s)(5 s)=75 m$

5 0

3 years ago

A steady beam of alpha particles (q = + 2e, mass m = 6.68 × 10-27 kg) traveling with constant kinetic energy 22 MeV carries a cu

cluponka [151]

Answer:

Explanation:

q = 2e = 3.2 x 10^-19 C

mass, m = 6.68 x 10^-27 kg

Kinetic energy, K = 22 MeV

Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A

(a) time, t = 2.8 s

Let N be the alpha particles strike the surface.

N x 2e = q

N x 3.2 x 10^-19 = i t

N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8

N = 2.36 x 10^12

(b) Length, L = 16 cm = 0.16 m

Let N be the alpha particles

K = 0.5 x mv²

22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²

v² = 1.054 x 10^15

v = 3.25 x 10^7 m/s

So, N x 2e = i x t

N x 2e = i x L / v

N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)

N = 4153.85

Kinetic energy = Potential energy

K = q x V

22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V

V = 1.17 x 10^7 V

5 0

3 years ago