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2 years ago

9

3. When the procedure is repeated with a third line how will it distinguish whether the location of the center of gravity is acc

urate or not?

1 answer:

Law Incorporation [45]2 years ago

4 0

Well this is question is easy. I mean i’m the one to say the least, The answer is.. SIKE L bozo imagine not knowing the answer

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As a bat flies toward a wall at a speed of 6.0 m/s, the bat emits an ultrasonic sound wave with frequency 30.0 kHz. What frequen

Leto [7]

Answer:

$f_b = 29.98 Hz$

Explanation:

speed of bat = 6 m/s

sound wave frequency emitted by bat = 30.0 kHz

as we know,

speed of sound (c)= 343 m/s

$f_w = f(\dfrac{c+v_r}{c+v_s})$

$f_w = 30(\dfrac{343+0}{343+6})$

$f_w = 29.48 Hz$

now frequency received by bat is equal to

$f_b = f(\dfrac{c+v_r}{c+v_s})$

$f_b = 29.48(\dfrac{343+6}{343+0})$

$f_b = 29.98 Hz$

hence the frequency hear by bat will be 29.98 Hz

7 0

3 years ago

If you can simply pour sand into a cup then why is it not a liquid?

sleet_krkn [62]

If you, for example, poured it onto a wide cup with a volume equal to the total volume of the sand particles, the sand would not spread out to fill the container but would bunch up together in the middle.

6 0

3 years ago

Read 2 more answers

The unit of electric potential or electromotive force is the

Natalka [10]

That would be called VOLT

4 0

3 years ago

In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles

alexandr402 [8]

Answer:

Explanation:

We shall apply conservation of mechanical energy

kinetic energy of alpha particle is converted into electric potential energy.

1/2 mv² = k q₁q₂/d , d is closest distance

d = 2kq₁q₂ / mv²

= 2 x 9 x 10⁹ x 79e x 2e / 4mv²

= 1422 x2x (1.6 x 10⁻¹⁹)² x 10⁹ /4x 1.67 x 10⁻²⁷ x (1.5 x 10⁷)²

= 3640.32 x 10⁻²⁹ /2x 3.7575 x 10⁻¹³

= 484.4 x 10⁻¹⁶

=48.4 x 10⁻¹⁵ m

8 0

3 years ago

Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and

tatuchka [14]

Answer:

solution:

dT/dx =T2-T1/L

&

q_x = -k*(dT/dx)

<u>Case (1) </u>

dT/dx= (-20-50)/0.35==> -280 K/m

q_x =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

q_x =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

q_x =-50*(80)*10^3==>-4 kW

Case (3)

q_x =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

6 0

3 years ago