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Arisa [49]
3 years ago
6

In an internal combustion engine, the gas vapor/air mixture enters the cylinder during the _____ stroke. intake compression powe

r exhaust
Physics
1 answer:
grigory [225]3 years ago
3 0
The cylinder takes in the gas/air mixture during the intake stroke.
The cylinder compresses the gas/air mixture during the compression stroke.
The gas/air mixture burns and generates power during the power stroke.
The cylinder forces the gas/air mixture out during the exhaust stroke.
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A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of
puteri [66]

The frequency and amplitude of the SHM beam is 0.8 Hz and 0.098 m. The frequency of the SHM wave when gravel falls is 0.8 Hz and the amplitude of subsequent SHM beam is 0.4m.

(a) Mass of the spring = 225 Kg

Mass of the sack = 175 Kg

Amplitude of the beam = 40 cm = 0.40 m

Frequency of the beam = F = 0.60 cycles/s

The formula for frequency of oscillation =

= f = (1/2π) X √(k/m)

where, k = 2π²F²m

= k = 2 X (3.14)² X 0.6² X (225 + 175)

= k = 5685.37 N/m

Strength of the spring before gravels fall = x =

= x = mg / k

= x = [ (225 + 175 ) X 9.8 ] / 5685.37

= x = 0.689 m

But, the spring is stretched by the distance of x' which is expressed as,

= X = x - x'

= X = 0.689 - 0.40

= X = 0.289 m

Now, since we know that the gravel falls, thus frequency = f =

= f = (1/2π) X √(k/m)

= f= (1/ 2 X 3.14) X √ 5685.37 / 225

= f = 0.8 Hz

(b) Assuming that the spring is stretched, x = mg/k =

= x = (225 X 9.8) / 5685.37

= x = 0.3878 m

Thus, the amplitude of the sack = A = 0.3878 - 0.289

= A = 0.098 m

(c) If the gravel falls, the speed is maximum hence speed = s =

= s = A X √(k/m)

= s = 0.4 X √(5685.37/400)

= s = 1.508 m/s

The frequency = f' =

= f' = (1/2π) X √(k/m)

= f' = (1/2 X 2.14) X √(5685.37/225)

= f' = 0.8 Hz

(d) New amplitude = A' =

= A' = 0.38 + 0.038   (after calculating the new distance)

= A' = 0.4 m

To know more about Spring:

brainly.com/question/15850235

#SPJ4

8 0
2 years ago
Suppose you take two non-zero displacements represented by vectors A & B.The magnitude of A is 5 m and the magnitude of B is
olga55 [171]

Answer:

a. When the total displacement is -(A + B)

b. A + B = 1 m or -(A + B) = -11 m

c. 0 m

Explanation:

a. Under what circumstances can you end up back at your starting point?

If we have the displacement A and displacement B. The total displacement is A + B. We would end up at the starting point if we take a displacement -(A + B) from point B

b. What is the magnitude of the largest displacement you can end up from the starting point?

The maximum displacement we can obtain is when A and B are in the same direction. So A + B = 5 m + 6 m = 11 m or -A - B = -(A + B) = -11 m.

c. When A and B are perpendicular, what is the component of B in the direction of A?

Since A is perpendicular to B, the angle between A and B is 90°

So the component of B in A,s direction is Bcos90° = B × 0 = 0 m

8 0
3 years ago
The 9-m boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point Clocated on the vertica
Vanyuwa [196]

Answer:

27000 Nm

Explanation:

The boom end at A is fixed and end at B is subjected to a 3kN force. Boom AB has length of 9m. The moment about A is the product of force 3kN (3000 N) at B and the moment arm of 9m

M = FL = 3000 * 9 = 27000 Nm

So the moment about A is 27000 Nm

7 0
3 years ago
A vertical spring (ignore its mass), whose spring stiffness constant is 880 n/m, is attached to a table and is compressed down 0
sladkih [1.3K]
When the spring is compressed, the potential energy stored in the spring is:
U= \frac{1}{2}kx^2
where k=880 N/m is the spring constant and x=0.170 m is the compression of the spring. By using these numbers, we get
U= \frac{1}{2}(880 N/m)(0.170 m)^2=12.7 J

When the spring is released with the ball over it, all the potential energy is converted into kinetic energy of the ball:
U=K= \frac{1}{2} mv^2
So by using m=0.300 kg, and re-arranging the formula, we can calculate the velocity of the ball:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 12.7 J}{0.300 kg} }=9.2 m/s
5 0
3 years ago
Describe how you determine whether an object is in motion
PolarNik [594]

To determine whether an object is in motion or not, you first
need to specify a reference point, because there's no such
thing as "real" motion, only motion relative to something.

Once you've named the reference point, you have to look at
the object at two different times.  Each time you look at it, you
measure its distance and direction from the reference point. 
If there's any difference in these measurements from one time
to the next, then the object has had average motion during the
period between the two observations.

That's the best you can do ... find average motion during some
period of time.  You can never definitely tell whether or not the
object ever stopped during that time.  But you can sneak up on
it by making the time period between the two observations shorter
and shorter.

6 0
3 years ago
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