Answer:

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



now frequency received by bat is equal to



hence the frequency hear by bat will be 29.98 Hz
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.
That would be called VOLT
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
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