initial speed of the racer is given as
after applied force the final speed is given as
now during this speed change the racer will cover total distance 185 m
so here we will use kinematics
now the force that chute will exert on the racer will be given as
B) here following is the strategy for solving it
1. first we used kinematics to find the acceleration of the car
2. then we used Newton's II law (F = ma) to find the force
Answer:
kinetic energy is 2, 4, 50 thirties, 2.45 into 10, raise to three Julie.
Explanation:
In the given question, we are told that what is the kinetic energy of mass M equals 0.1 kg bullet traveling at a velocity Velocity is given and 700 m/s. So we know that kinetic energy mm-hmm k equals one half m v squared. So this will be mass is given 0.1 and velocity is 700 so 700 square this is one half 0.1 in two 49 double zero, double zero. This is one-half into 49 double zero. So kinetic energy is 2, 4, 50 thirties, 2.45 into 10, raise to three Julie. This is kinetic energy. Thank you.
Kinetic energy = (1/2) (mass) x (speed)²
At 7.5 m/s, the object's KE is (1/2) (7.5) (7.5)² = 210.9375 joules
At 11.5 m/s, the object's KE is (1/2) (7.5) (11.5)² = 495.9375 joules
The additional energy needed to speed the object up from 7.5 m/s
to 11.5 m/s is (495.9375 - 210.9375) = <em>285 joules</em>.
That energy has to come from somewhere. Without friction, that's exactly
the amount of work that must be done to the object in order to raise its
speed by that much.
Answer:
The amount of current that must flow through the wire for it to be suspended against gravity by magnetic force = 6.125 A
Explanation:
Force on a wire carrying current in an electric field is given by
F = (B)(I)(L) sin θ
For this question,
The magnetic force must match the weight of the wire.
F = mg
mg = (B)(I)(L) sin θ
(m/L)g = (B)(I) sin θ
Mass per unit length = 75 g/m = 0.075 kg/m
B = magnetic field = 0.12 T
I = ?
g = acceleration due to gravity = 9.8 m/s
θ = angle between wire's current direction and magnetic field = 90°
0.075 × 9.8 = 0.12 × I sin 90°
I = 0.075 × 9.8/0.12 = 6.125 A
Answer:
Temperature of water leaving the radiator = 160°F
Explanation:
Heat released = (ṁcΔT)
Heat released = 20000 btu/hr = 5861.42 W
ṁ = mass flowrate = density × volumetric flow rate
Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³
ṁ = 1000 × 0.000126 = 0.126 kg/s
c = specific heat capacity for water = 4200 J/kg.K
H = ṁcΔT = 5861.42
ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C
And in change in temperature terms,
10°C= 18°F
11.08°C = 11.08 × 18/10 = 20°F
ΔT = T₁ - T₂
20 = 180 - T₂
T₂ = 160°F
