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2 years ago

To calculate lobe lift, you need to know the _______ diameter and the lobe height

2 answers:

5 0

The answer is the diameter of the cam shaft.

This is used to compute for the area of the circle so you can multiply it with the rocker ratio to get maximum amount of weight the valve can lift.

PtichkaEL [24]2 years ago

4 0

Answer: Journal

Explanation:

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A rock of mass m is thrown horizontally off a building from a height h. the speed of the rock as it leaves the thrower's hand at

Stells [14]

The correct answer is 3) $K_f = \frac{1}{2}mv_0^2 + mgh$ .

In fact, the total energy of the rock when it leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
 $E=U_i+K_i=mgh + \frac{1}{2}mv_0^2$
As the rock falls down, its height h from the ground decreases, eventually reaching zero just before hitting the ground. This means that U, the potential energy just before hitting the ground, is zero, and the total final energy is just kinetic energy:
 $E=K_f$ 
But for the law of conservation of energy, the total final energy must be equal to the tinitial energy, so E is always the same. Therefore, the final kinetic energy must be
 $K_f = mgh + \frac{1}{2}mv_0^2$

7 0

2 years ago

A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings

forsale [732]

Answer:v=2 m/s

Explanation:

Given

Length of string L=1.2 m

mass of pendulum m=0.25 kg

maximum inclination with vertical \theta =34

vertical Rise of Pendulum from its mean position is given by

$\Delta h=L(1-\cos \theta )$

Conserving Energy at top and bottom point

Potential Energy of sphere is converted into kinetic energy of sphere

$mgL(1-\cos \theta )=\frac{mv^2}{2}$

$v=\sqrt{2gL(1-\cos \theta )}$

$v=\sqrt{2\times 9.8\times 1.2(1-\cos 34)}$

$v=\sqrt{4.021}$

$v=2 m/s$

5 0

2 years ago

Trained dolphins are capable of a vertical leap of 7.0 m straight up from the surface of the water - an impressive feat. Suppose

dmitriy555 [2]

Answer:14 m

Explanation:

Given

Vertical jump make by the dolphin is given by $h=7\ m$

Suppose the dolphin jump with an initial velocity of $u$

so u is given by $u^2=2\cdot g\cdot h$

If dolphin launches at an angle $\theta$ then maximum horizontal range is given by

assuming the of Dolphin to be Projectile so range is given by

$R=\frac{u^2\sin 2\theta }{g}$

substitute the value of $u^2$

$R=\frac{2\times 9.8\times 7\sin 2\theta }{9.8}$

$R=2h\sin 2\theta$

Range will be maximum for $\theta =45^{\circ}$

thus $R_{max}=2\times 7\times 1=14\ m$

3 0

2 years ago

A 2.0-m-tall man is 5.0 m from the converging lens of a camera. His image appears on a detector that is 50 mm behind the lens. H

ladessa [460]

Answer:

20 cm

Explanation:

We can solve the problem by using the magnification equation:

$M=\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the size of the image

$h_o = 2.0 m$ is the height of the real object (the man)

$q=50 mm =0.050 m$ is the distance of the image from the lens

$p = 5.0 m$ is the distance of the object (the man) from the lens

Solving the formula for $h_i$ , we find

$h_i = -\frac{q}{p}h_o=-\frac{0.050 m}{5.0 m}(2.0 m)=-0.02 m = -20 cm$

And the negative sign means the image is inverted.

6 0

2 years ago

A 20.0 kg curling stone travels 30.0 m along the ice surface. If the frictional force is 10.0 N, the thermal energy produced is

evablogger [386]

The thermal energy is where the work of friction comes from. That is what stops it eventually. In this case a counter force of 10N is applied over the distance of 30.0m. The energy is given by Force*Distance. Here this is 300J. This friction work is the thermal energy.

4 0

2 years ago

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