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3 years ago

Match the particles with their characteristics,

Physics

2 answers:

In-s [12.5K]3 years ago

6 0

Positive charge=proton
Negative charge=electron
No charge/neutral=neutron

Pepsi [2]3 years ago

4 0

Answer:

Neuton has no charge =proton has positive charge >electron has negative charge.

Explanation:

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7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf

Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to 1 μV/m

Intrinsic impedance of air = 120 π Ω

Intrinsic impedance of water = $\dfrac{120\pi}{\epsilon_r}$

Using equation to solve the problem

$E(z) = E_0 e^{-\alpha\ z}$

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

$\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}$

$\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}$

$\alpha =21.07\ Np/m$

now ,

$1 \times 10^{-6} = 20 e^{-21.07\times z}$

$e^{21.07\times z}= 20\times 10^{6}$

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0

3 years ago

A ball is rolled uphill a distance of 12 meters before it slows, stops, and begins to roll back. The ball rolls downhill 20 mete

Sedaia [141]

The ball rolled a distance of
d = 12m + 20m.
But the change of position is
x = + 12m - 20m

5 0

3 years ago

The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for

pshichka [43]

Answer:

d = 69 .57 meter

Explanation:

First case

Speed of car ( v ) = 20.5 mi/h = 9.164 M/S

distance ( d ) = 11.6 meter ( m = mass of the car )

Work done = 0.5 m v² = 0.5 * 9.164² * m J = 41.99 m J

Force = ( workdone /distance ) = ( 41.99 m / 11.6 ) = 3.619 m N

Second case

v = 50.2 mi/h = 22.44135 m/s

d = ?

Work done = 0.5 * 22.44² * m J = 251.7768 * m J

Since the braking force remains the same .

3.619 m = ( 251.7768 m / d )

d = 69 .57 meter

7 0

3 years ago

A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the o

katen-ka-za [31]

Answer:

Explanation:

Step one:

<u>Given </u>

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

<u>Required</u>

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

4 0

3 years ago

In a ballistics test, a 52g bullet hits a sand bag and stops after moving 1.34 m. If the initial bullat

olya-2409 [2.1K]

Answer:

Friction force on the bullet is 58.7 N opposite to its velocity

Explanation:

As we know that initial speed of the bullet is 55 m/s

after travelling into the sand bag by distance d = 1.34 m it comes to rest

so final speed

$v_f = 0$

now we can use kinematics top find the acceleration of the bullet

$v_f^2 - v_i^2 = 2 a d$

so we have

$0 - 55^2 = 2(a)(1.34)$

$a = -1128.7 m/s^2$

now by Newton's II law we know that

$F = ma$

so we have

$F = (0.052)(-1128.7)$

$F = -58.7 N$

8 0

3 years ago