The work done is 
Explanation:
The work done by a force on an object is given by:
where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and of the displacement
In this problem, we have
F = 1080 N is the force applied on the car
d = 218 m is the displacement of the car
And assuming the force is applied parallel to the motion of the car, 
, and so the work done is
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Answer:
Distance, d = 192 meters
Explanation:
We have,
Initial velocity of an object is 10 m/s
Acceleration of the object is 3.5 m/s²
Time, t = 8 s
We need to find the distance travelled by the object during that time. Second equation of motion gives the distance travelled by the object. It is given by :
So, the distance travelled by the object is 192 meters.
Answer:
Psm = 30.66 [Psig]
Explanation:
To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.
P * v = R * T
where:
P = pressure [Pa]
v = specific volume [m^3/kg]
R = gas constant for air = 0.287 [kJ/kg*K]
T = temperature [K]
<u>For the initial state</u>
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P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)
T1 = -2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)
Therefore we can calculate the specific volume:
v1 = R*T1 / P1
v1 = (0.287 * 270.4) / 266.8
v1 = 0.29 [m^3/kg]
As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.
V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.
T2 = 43 + 273 = 316 [K]
P2 = R*T2 / V2
P2 = (0.287 * 316) / 0.29
P2 = 312.73 [kPa]
Now calculating the manometric pressure
Psm = 312.73 -101.325 = 211.4 [kPa]
And converting this value to Psig
Psm = 30.66 [Psig]
Period and frequency are mutual reciprocals.
Period = 1 / frequency .
Frequency = 1 / period
(Frequency) x (Period) = 1
Answer:
r = 0.303m
= 30.3cm
Explanation:
Given that,
The number of electrons transferred from one sphere to the other,
n =
1
×10
¹³e
le
c
t
r
o
n
s
The electrostatic potential energy between the spheres,
U
=
−
0.061
J
The charge on an electron,
q
=
−
1.6
×
10
⁻¹⁹C
The coulomb constant,
K
=
8.98755
×
10
⁹
N
⋅
m
²
/
C
2²
Due to the transfer of electrons, both spheres become equally and oppositely.
The charge gained by the sphere due to the excess of the electron is:
q
₁ =
n
q
=
1
×10
¹³ * −
1.6
×
10
⁻¹⁹
= -1.6 × 10⁻⁶C
The charge left on the first sphere is =
q
₂ = -q₁ = 1.6 × 10⁻⁶C
The electric potential energy between two point charges is given by the following equation:
U
=
K
q
₁q
₂/r
q
₁ and q
₂ are the two charges.
r is the distance between the charge and the point.
K = 8.98755 × 10
⁹
N
⋅
m
²
/
C
²
we have:
-0.061 = (8.98755 × 10
⁹ * (-1.6 × 10⁻⁶)²) / r
r = (18.41 × 10
⁻³) / 0.061
r = 0.303m
= 30.3cm
