F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
5.5 s
Explanation:
The time it takes for the ball to reach its maximum height can be calculated using

since
at the top of its trajectory. Plugging in the numbers,

Answer:
Sarah is right
Explanation:
This is an exercise that differentiates between scalars and vectors.
A scalar is a number, instead a vector is a number that represents the module in addition to direction and sense.
In this case, the distance (scalar) traveled is a number, which is why it is worth 1500m, but the displacement is a vector and since the point where it leaves is the same point where the vector's modulus arrives is zero, so the DISPLACEMENT VECTOR is zero
consequently Sarah is right
Answer:
a) 2.063*10^-4
b) 1.75*10^-4
Explanation:
Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper </em>
<em> wire: </em>
L= 2.00 m
From Table Copper Resistivity
= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

=0.0165Ω
The Potential difference across the copper wire is:
V=IR
=2.063*10^-4
b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m
The Resistance of the Silver wire is:

=0.014Ω
The Potential difference across the Silver wire is:
V=IR
=1.75*10^-4