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3 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]3 years ago

4 0

Answer: Journal

Explanation:

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A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north

Sedaia [141]

Answer:

Magnitude of displacement = 2.07 km

Magnitude of average velocity = 1.17 kmph

Explanation:

Let east represent positive x axis and north represent positive y axis.

A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north of west.

1.93 km due wast

s ₁ = 1.93 i km

1.03 km due south

s₂ = -1.03 j km

3.84 km in a direction 52.8 ° north of west

s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km

Total displacement

s = s ₁+ s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j

Magnitude of displacement, $=\sqrt{(-0.39)^2+2.03^2}=2.07km$

Time taken = 1.771 hour

Magnitude of average velocity, $=\frac{2.07}{1.771}=1.17km/hr$

7 0

3 years ago

Sound waves need a medium to move through in order to keep moving. Through which of the following would sound travel the slowest

Troyanec [42]

Answer:A air

Explanation:Of the three mediums (gas, liquid, and solid) sound waves travel the slowest through gases, faster through liquids, and fastest through solids. Temperature also affects the speed of sound.

Hope this helps you out ツ

8 0

3 years ago

Read 2 more answers

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

4 years ago

Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1.05 m.At what minimum angle relative to the c

kiruha [24]

Answer:

15.19°, 31.61°, 51.84°

Explanation:

We need to fin the angle for m=1,2,3

We know that the expression for wavelenght is,

$\lambda = \frac{c}{f}$

Substituting,

$\lambda = \frac{344}{1250}$

$\lambda = 0.2752m$

Once we have the wavelenght we can find the angle by the equation of the single slit difraction,

$sin\theta = \frac{m \lambda}{W}$

Where,

W is the width

m is the integer

$\lambda$ the wavelenght

Re-arrange the expression,

$\theta = sin^{-1} \frac{m\lambda}{W}$

For m=1,

$\theta = sin^{-1} \frac{1 (0.2752)}{1.05}= 15.19\°$

For m=2,

$\theta = sin^{-1} \frac{2 (0.2752)}{1.05}= 31.61\°$

For m=3,

$\theta = sin^{-1} \frac{3 (0.2752)}{1.05}= 51.84\°$

<em>The angle of diffraction is directly proportional to the size of the wavelength.</em>

5 0

3 years ago

A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed

Vladimir [108]

Answer:

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

Explanation:

A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?

It decreases in speed on its way down and increases in speed on its way down.

it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center

.It increases in speed on his way down because its under the influence of gravity

from newton's equation of motion we can check by

using V^2=u^2+2as

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

5 0

3 years ago