Answer:3.4 seconds
Explanation:
Initial velocity(u)=0
acceleration=34.5m/s^2
Height(h)=200m
Time =t
h=u x t - (gxt^2)/2
200=0xt+(34.5xt^2)/2
200=34.5t^2/2
Cross multiply
200x2=34.5t^2
400=34.5t^2
Divide both sides by 34.5
400/34.5=34.5t^2/34.5
11.59=t^2
t^2=11.59
Take them square root of both sides
t=√(11.59)
t=3.4 seconds
Answer:
27.44 J
Explanation:
We can find the energy at the top of the slide by using the potential energy equation:
At the top of the slide, the swimmer has 0 kinetic energy and maximum potential energy.
The swimmer's mass is given as 7.00 kg.
The acceleration due to gravity is 9.8 m/s².
The (vertical) height of the water slide is 0.40 m.
Substitute these values into the potential energy equation:
- PE = (7.00)(9.8)(0.40)
- PE = 27.44
Since there is 0 kinetic energy at the top of the slide, the total energy present is the swimmer's potential energy.
Therefore, the answer is 27.44 J of energy when the swimmer is at the top of the slide.
Answer:
10500 J/kg/*C
Explanation:
Quantity of heat required=mass of substance x specific heat capacity x change in temperature
Quantity of heat required=0.25 x 4200 x [30-20]
Quantity of heat required=0.25 x 4200 x 10
Quantity of heat required=10500 J/kg/*C
Answer:=14,160 kJ
Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below
Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase
C MPa m^3/kg kJ/kg kJ/kg
1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture
2 260 4.689 0.0422 2599 2797 1 Saturated Vapor
The mass initially contained in the tank is m1 = V/v1
m1 =0.85 m^3 /0.02993 m^3 /kg
= 28.4 kg
The mass finally contained in the tank is
m2 =V2/v
= 0.85 m^3 /0.0422 m^3 /kg
= 20.14 kg
The heat transfer is then
Qcv = m2u2 − m1u1 − he(m2 − m1)
Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ
Answer: cleft grafting, inlay grafting, four-flap grafting, and whip grafting.
Explanation:
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