Answer:
The distance travel before stopping is 1.84 m
Explanation:
Given :
coefficient of kinetic friction 
Zak's speed 
Gravitational acceleration 
Work done by frictional force is given by,
m
Therefore, the distance travel before stopping is 1.84 m
ANSWER:
F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.
STEP-BY-STEP EXPLANATION:
F(h) is Horizontal Force = 200 N
V is Speed = 2.4 m/s
The total weight increase by 42%
coefficient of rolling friction decrease by 19%
Since the velocity is constant so acceleration is zero; a=0
Now the horizontal force required to move the pickup is equal to the frictional force.
F(h) = F(f)
F(h) = mg* u
m is mass
g is gravitational acceleration = 9.8 m/s^2
200 = mg*u
Since weight increases by 42% and friction coefficient decreases by 19%
New weight = 1+0.42 = 1.42 = (1.42*m*g)
New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u
F(h) = (0.81μ) (1.42 m g)
= (0.81) (1.42) (μ m g)
= (0.81) (1.42) (200)
= 230 N
Answer:
Explanation:
Given:
- mass of rocket,
- time of observation,
- mass lost by the rocket by expulsion of air,
- velocity of air,
<u>Now the momentum of air will be equal to the momentum of rocket in the opposite direction: </u>(Using the theory of elastic collision)
The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.
Therefore, option A is correct option.
Given,
Mass m = 14g
Volume= 3.5L
Temperature T= 75+273 = 348 K
Molar mass of CO = 28g/mol
Universal gas constant R= 0.082057L
Number of moles in 14 g of CO is
n= mass/ molar mass
= 14/28
= 0.5 mol
As we know that
PV= nRT
P × 3.5 = 0.5 × 0.082057 × 348
P × 3.5 = 14.277
P = 14.277/3.5
P = 4.0794 atm
P = 4.1 atm.
Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.
learn more about pressure:
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