Answer:
Gas mileage is 17.29
Explanation:
Given data:
The total number of the model is 35
The total size of the engine is 5 ltr
The regression model is given as
From the information given in question we have
Regression equation is : model- mpg
Therefore for engine capacity of 5 liters;
Gas mileage
Gas mileage is 17.29
Answer:16.
Explanation:
We have that for the Question it can be said that the magnitude of the force exerted by the horizontal rope on her arms and the ratio of the Force to the weight is
From the question we are told
A 64.0-kg ice skater is moving at 4.04 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.890 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. kN (b) Compare this force with her weight. F-rope W =
Generally the equation for the force applied is mathematically given as
F=1150.561N
b)
Generally the equation for the Weight is mathematically given as
W=mg
Therefore
W=64*9.81
W=627.84N
Therefore
The Force to weight ratio is
F/W=1.8325
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The force exerted by the rope on the skater's arms as she moves in a circular path is 1.167 kN. This force is about 1.860 times her weight, which is 627.2 N.
The skater is experiencing centripetal force exerted by the rope, which causes her to move in a circular path. The magnitude F of this force can be calculated using the formula F = mv²/r, where m is the skater's mass (64.0 kg), v is her velocity (4.04 m/s), and r is the radius of her circular path (0.890 m).
By substituting the given numbers into this formula, we get: F = (64.0 kg)(4.04 m/s)² / 0.890 m = 1166.67 N. In kilonewtons, this force is 1.167 kN.
To compare this force with her weight, we can calculate the weight (W) using the formula W = mg, where g is the acceleration due to gravity (around 9.8 m/s²). Substituting the given mass into this formula gives us: W = (64.0 kg)(9.8 m/s²) = 627.2 N.
Comparing these two forces shows that the force exerted by the rope on her arms is about 1.860 times her weight.
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Answer:
If the acceleration is constant, the movements equations are:
a(t) = A.
for the velocity we can integrate over time:
v(t) = A*t + v0
where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:
Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.
Explanation:
e(t) = 9t² − 6t + 3
The velocity is the first derivative:
e'(t) = 18t − 6
The acceleration is the second derivative:
e"(t) = 18
Answer:
b
Explanation:
Answer:
D = 58 cm
Explanation:
Given that,
Focal length of the objective lens,
Focal length of the eye piece,
We need to find how many cm apart should the lenses be placed. Let d be the distance between lenses. It is equal to the sum of focal lengths of objective lens and eye-piece
D = 48 cm + 10 cm
= 58 cm
Hence, the object is placed at a distance of 58 cm.
In an astronomical telescope, the lenses should be placed at a distance equal to the sum of their focal lengths. In this case, this distance would be 58 cm.
In an astronomical telescope, the distance between the objective lens and the eyepiece should be equal to the sum of their focal lengths for the telescope to produce clear and sharp images. Here, the focal length of the objective lens is 48 cm and the focal length of the eyepiece is 10 cm.
Therefore, calculating: Objective lens focal length + Eyepiece focal length = 48 cm (objective) + 10 cm (eyepiece) = 58 cm
This means that the objective lens and the eyepiece should be approximately 58 centimeters apart.
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