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3 years ago

Which of the following would most likely utilize a particulate control device?

Physics

2 answers:

Alexandra [31]3 years ago

3 0

Answer:

Explanation:

Correct answer: D). All of the above The particulate control device is a device that is used for the separation of the material that is of a bigger size as compared to filter. In the factory smokestack, the particulate control device filters the particulate matter that is known to cause air pollution. In the vacuum cleaner, a fine dust particle is filtered out. In the laundry dryer, the water and dirt molecule is allowed to pass through it. Hence, all the options are correct.so therefore your answer would be all

Ierofanga [76]3 years ago

3 0

Answer: all of the above

Explanation:

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Which data set has the largest range? A. 55, 57, 59, 60, 61, 49, 48 B. 21, 25, 14, 16, 29, 22, 20 C. 12, 15, 16, 19, 18, 15, 27

Simora [160]

Data D has the largest range.

Data A: 61-48=13
Data B: 29-14=15
Data C:27-12=15
Data D:54-31=23

Therefore, Data D has the largest range.

3 0

3 years ago

Pls help me with this question

Arisa [49]

Answer:

dam 15 marks for that question that's ez marks there

4 0

3 years ago

A dragster starts with zero velocity and completes a 404.5 m (0.2528 mile) run in 4.922 s. If the car had a constant acceleratio

KIM [24]

Answer:

a. a=33.34ms⁻², V=164.4m/s

Explanation:

Since the dragster started with zero velocity, de determine the acceleration using of the equations of motion.

Below are the data given

Distance, s=404.5m,

time taken,t=4.922secs

Using the equation

S=ut+1/2at²

where u is the initial velocity and u=0

Making the acceleration the subject of the formula, we arrive at

a=2s/t²

a=(2*404.5)/4.922²

a=33.34ms⁻².

To determine the velocity, we use

V=u+at

V=0+33.34ms⁻² *4.922sec

V=164.4m/s

5 0

3 years ago

Read 2 more answers

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

$d=\frac{1,280,000,000km}{37,000}=34,594.59km$

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

$v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}$

hence, the speed of the Hubble is approximately 384km/min

5 0

3 years ago

You are given a vector in the xy plane that has a magnitude of 90.0 units and a y component of -61.0 units. A) Assuming the x co

kvasek [131]

Answer:

A) magnitude = sqrrt(147.1^2 + 61^2) = 159.2 units

B)22.5° (clockwise form -ve X axis)

Explanation:

given vector = V1 = x i - 60 j

magnitude of V1 = 90

x - component can be found out by resultant formula

90^2 = x^2 + (-60)^2

x = 67.08 = 67.1 units (3sf)

FOR THE VECTOR 80 UNITS IN -VE X DIRECTION

The X component is -80-------(1)

The Y component is 0 ---------(2)

For the x- component of new added vector:

(1)----------- x + 67.1 = -80

x = -147.1 = -147.1

For the y- component of new added vector:

(2)---------- y - 61 = 0

y = 61.0 (3sf)

the new added vector is = -147.1 i + 61 j

magnitude = sqrrt(147.1^2 + 61^2) = 159.2 units

direction = arctan (61 / 147.1)

= 22.5° (clockwise form -ve X axis)

4 0

3 years ago