Answer:
The equation that we can used to find the maximum temperatures at which nitrogen is a liquid, X is;
Explanation:
To solve the question, we note that the temperature of nitrogen has
Median value of the nitrogen temperature range = -333.22 Fahrenheit
Range of the temperatures values at which nitrogen is a liquid from the median = 12.78 Fahrenheit
Therefore, maximum temperature at which nitrogen exists as a liquid is
12.78 + -333.22 = 320.44 Fahrenheit
Also the minimum temperature at which nitrogen exists as a liquid is given as
- 12.78 + -333.22 = -346 Fahrenheit
The equation that can be used to find the maximum temperatures at which nitrogen is a liquid, X is therefore,
Answer:
If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Explanation:
R₁ = Resistance of first resistor
R₂ = Resistance of second resistor
V = Voltage of battery = 12 V
I = Current = 0.33 A (series)
I = Current = 1.6 A (parallel)
In series
In parallel
Solving the above quadratic equation
∴ If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Answer:
The force applied on the big piston is 1306.67 N
Explanation:
Given;
force applied on small piston, F₁ = 200 N
diameter of the small piston, d₁ = 4.37 cm
radius of the small piston, r₁ = d₁/2 = 2.185 cm
Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²
Area of the big piston, A₂ = 98 cm²
The pressure of the piston is given by;
Where;
F₂ is the force on big piston
Therefore, the force applied on the big piston is 1306.67 N
Answer:
The net force on the stump is 1000 N.
Explanation:
Given that,
Force 1 acting on the truck, (due north)
Force 2 acting on the truck, (due west)
We need to find the net force on the stump. We know that force is a vector quantity. The net force on the stump is given by the the resultant force. It is given by :
F = 1000 N
So, the net force on the stump is 1000 N. Hence, this is the required solution.
Answer:
1) 64.2 mi/h
2) 3.31 seconds
3) 47.5 m
4) 5.26 seconds
Explanation:
t = Time taken = 2.5 s
u = Initial velocity = 0 m/s
v = Final velocity = 21.7 m/s
s = Displacement
a = Acceleration
1) Top speed = 28.7 m/s
1 mile = 1609.344 m
1 hour = 60×60 seconds
Top speed of the cheetah is 64.2 mi/h
Equation of motion
Acceleration of the cheetah is 8.68 m/s²
2)
It takes a cheetah 3.31 seconds to reach its top speed.
3)
It travels 47.5 m in that time
4) When s = 120 m
The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds