Answer:
the rate of acceleration of the train is 4 m/s²
Explanation:
Given;
initial velocity of the train, u = 10 m/s
change in time of motion, dt = 5 s
final velocity of the train, v = 30 m/s
The rate of acceleration of the train is calculated as;
Therefore, the rate of acceleration of the train is 4 m/s²
As we know that P=IV so P = (0.5A)(4.5V)
P= 2.25 watt
Answer:32km/hr
Explanation:
Speed=distance/time
Speed=60/2 =30km/hr speed=68/2=34km/he
Average speed=(30+34)/2 =64/2=32km/hr
Answer:
A. -5488J
B. 273.8J
C. 372.44N
Explanation:
Given:
m = 40kg
h = 14 m
v= 3.7 m/s
Part(a)
The change in the potential energy of the bear Earth system during the slide
AU = -mgh = -40(9.8) (14) = -5488 J
Part(b)
The kinetic energy of the bear just before hitting the ground is
Ks 1/2 mV^2= (40)(3.7)2 = 547.6 /2 = 273.8J
Part(c)
The change in the thermal energy of the system due to friction is
AEth = fxh=-(AK +AU) = 5488– 273.8 = 5214.2 J
The average frictional force that acts on the sliding bear is
F = Eth / 14= 5214.2/14 =372.44N
Answer:
Explanation:
Angle of dip = 56° , magnetic field strength = 50μT
Vertical component = 50 x sin 56 = 41.45 μT
Horizontal component = 50 cos 56 = 27.96μT
New field is added in vertical downwards direction to increase the vertical component so as to increase the angle of dip . Let this field be B
total vertical field = B + 41.45
Horizontal component = 27.96
dip angle be θ
tanθ = vertical component / horizontal component
tan62 = B + 41.45 / 27.96
1.88 = B + 41.45 / 27.96
52.58 = B + 41.45
B = 11.13 μT
Since magnetic field has to be added , current should be clockwise when looked from above.
