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geniusboy [140]
3 years ago
9

The acceleration of an object in uniform circular motion always points toward the center of the circle. Question 2 options: True

False
Physics
1 answer:
vitfil [10]3 years ago
7 0

Answer:

The answer is true. Hope this helps!

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Which of the following statements is true?
JulsSmile [24]
The answer would be #3
7 0
3 years ago
Read 2 more answers
acontainer is filled whith mercury to alevel of 10m whit water to alevel of 8m and whit oil to alevel of 5m the densities oil ,w
Alborosie

Answer:

1450.4 KNm^{2}

Explanation:

Pressure = ρhg

where: ρ is the density of the liquid, h is the height and g the force of gravity.

Total pressure exerted by the liquids at the base = Pressure of oil + Pressure of water + Pressure of mercury

So that,

i. Pressure of oil = ρhg

(ρ = 0.8 g/cm³ = 800 kg/m³)

                        = 800 x 5 x 9.8

                        = 39200

Pressure of oil = 39200 Nm^{2}

ii. Pressure of water = ρhg

(ρ = 1 g/cm³ = 1000 kg/m³)

                                      = 1000 x 8 x 9.8

                                     = 78400

Pressure of water = 78400 Nm^{2}

ii. Pressure of mercury = ρhg

(ρ = 13.6 g/cm³ = 13600 kg/m³)

                      = 13600 x 10 x 9.8

                      = 1332800

Pressure of mercury = 1332800 Nm^{2}

So that,

Total pressure exerted by the liquids at the base = 39200 + 78400 + 1332800

                                               = 1450400

                                               = 1450.4 KNm^{2}

Total pressure exerted by the liquids at the base is 1450.4 KNm^{2}.

8 0
3 years ago
A person walks 100m in 5
RUDIKE [14]
Answer:
20.35m per minute
Or
1,221.43 per hour





100/5=20
200/7= 28.57
50/4= 12.5

20+28.57+12.5= 61.07

61.07/3= 20.35m per minute

Or

20.35 x 60= 1,221.43m per hour
6 0
3 years ago
Find the equivalent resistance of this circuit
Phantasy [73]

Answer:

Req = 564 Ω

Explanation:

The equivalent resistance between R1 and R2:

1/R =1/R1 + 1/R2

1/R =1/960 + 1/640

1/R = 1/384

R = 384

Now, the equivalent resistance between R and R3:

Req = 384 + 180

Req = 564 Ω

8 0
3 years ago
While using a digital radiography system, suppose a radiographer uses exposure factors of 10 mAs and 70 kVp with an 8:1 grid for
Nimfa-mama [501]

Answer:

b. 12.5 mAs, 70 kVp

Explanation:

The given parameter are;

The initial exposure factors := 10 mAs and 70 kVp

The initial Grid Ratio, G.R.₁ = 8:1

The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂  = 12:1

Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;

The GCF for G.R. 8:1 = 4

The GCF for G.R. 12:1 = 5

Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;

mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs

Therefore the new exposure factors are;

12.5 mAs, 70 kVp

5 0
3 years ago
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