When running a 100 meter race Wyatt reaches his maximum speed when he is 40 meters from the starting line, and 7 seconds have elapsed since the start of the race. Wyatt continues at this max speed for the rest of the race and is 85 meters from the starting line 12 seconds after the start of the race. What is Wyatt's max speed
Answers
Answer:
9 m/s
Explanation:
Wyatt maintains the maximum speed for the rest of the race. This motion begins when his displacement is 40 m and the time is 7 s. At time 12 s, his displacement is 85 m. Because this motion is constant-velocity, the maximum speed is given by
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Answers
Answer:
-847.2J
Explanation:
First find the acceleration from v^2= u^2 + 2as
v= 2.5 m/s
u= 1.3 m/s
a???
s=6.00
a= v^2-u^2/2s
a= (2.5)^2-(1.3)^2/2× 6
a= 0.38ms^-2
From Newtons second law:
(Force applied cos Θ) - (Frictional force) = ma
Frictional force = ma- (Force applied cos Θ)
Frictional force= (18.8×0.38) - (165 cos 26°)
Frictional force= 7.144- 148.3= -141.2N
Therefore,
Work done by friction = Frictional force × distance covered
= -141.2N × 6= -847.2J
Answer:
W = –847J
Explanation:
Given m = 18.8kg, F = 165N, θ = -26° (below the horizontal, s = 6.0m, u = 1.30m/s and v = 2.50m/s
In this problem, two forces act on the chair; the forward force F and the frictional force f. We would apply newton's second law to find the frictional force f after which we can calculate the workdone by the frictional force f×s.
But for us to apply newton's second law, we need to know the acceleration of the chair cause by the net force.
From constant acceleration motion equations
v² = u² + 2as
2.5² = 1.30² + 2a×6
6.25 = 1.69 +12a
12a = 6.25 – 1.69
12a = 4.56a
a = 4.56/12
a = 0.38m/s
By newton's second law the net sum of forces equals m×a
The force F has horizontal and vertical and components. It is the horizontal component of this force that pushes the chair against friction.
Fx and f are oppositely directed.
So
Fx – f =ma
165cos(-26) – f = 18.8×0.38
148.3 – f = 7.14
f = 148.3 – 7.14
f = 141.2N
Workdone = -fs = –141.2×6.00 = –847J
W = –847J
Work is negative because it is done by a force acting on the chair in a direction opposite (antiparallel) to that of the intended motion.
Answers
Answer:
Explanation:
Magnetic moment of current carrying loop
= current x area
= 2 x π x .1²
M = .0628 unit . it is in j direction so vecor form of it
M = .0628 j
Magnetic field B = 3i + 4 j
Energy
= - M.B
- .0628 j . ( 3i + 4 j )
= - .2512 J
Answers
Answer:
Volume of gasoline spills out is 0.943 L.
Explanation:
Volumetric expansion of both gasoline and steel tank is :
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We know expansion due to temperature change is :
For gasoline:
Similarly for Steel tank:
.
Now, volume of gasoline spills out is equal to difference between expansion in volume.
Answers
Answer: 114.4 GJ
Explanation:
Heat loss Q=U×A×ΔT
Heat loss of size A is determined by the U value of materials and the difference in temperature.
From 10.9cm from the ice
50m= 5000cm
A= 5000×5000
Q== (10.9) (5000) (5000)(4.184)(1×4 + 80)
Q = 95,771,760,000J
Q≈ 95.8 GJ
Linear gradient from the bottom of the pond to the ice:
Q = (89.1)(5000)(5000)(4.184)(1*2)
Q = 18,639,720,000J
Q ≈ 18.6 GJ
Total heat loss:
Q= 95.8GJ + 18.6GJ
Q= 114.4 GJ
Answers
Answer:
29.5 m/s
Explanation:
Volumetric flowrate = (average velocity of flow) × (cross sectional area)
Volumetric flowrate = 0.111 liters/s = 0.000111 m³/s
Cross sectional Area of flow = πr²
Diameter = 0.00579 m,
Radius, r = d/2 = 0.002895 m
A = π(0.002895)² = 0.0000037629 m²
Velocity of flow = (volumetric flow rate)/(cross sectional Area of flow)
v = 0.000111/0.0000037629
v = 29.5 m/s
Given Information:
diameter of the nozzle = d = 5.79 mm = 0.00579 m
flow rate = 0.111 liters/sec
Required Information:
Velocity = v = ?
Answer:
Velocity = 4.21 m/s
Explanation:
As we know flow rate is given by
Flow rate = Velocity*Area of nozzle
Where
Area of nozzle = πr²
where
r = d/2
r = 0.00579/2
r = 0.002895 m
Area of nozzle = πr²
Area of nozzle = π(0.002895)²
Area of nozzle = 2.6329x10⁻⁵ m²
Velocity = Flow rate/area of nozzle
Divide the litters/s by 1000 to convert into m³/s
0.111/1000 = 1.11x10⁻⁴ m³/s
Velocity = 1.11x10⁻⁴/2.6329x10⁻⁵
Velocity = 4.21 m/s
Therefore, the water exit the nozzle at a speed of 4.21 m/s
Answers
Answer:
Explanation:
The equilibrium position of the sub is at the surface of the lake