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3 years ago

8

The point (8,15) lies on a circle centered at (0,0). Where does the circle intersect the -axis? Where does the circle intersect

the y-axis? Explain your answer.

1 answer:

sergey [27]3 years ago

4 0

Answer:

lee know

Step-by-step explanation:

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Can someone help me with this question

Olin [163]

The answer is 65 degrees because a triangle is 180 degrees in total so if you add 53 and 63 you get 115 and subtract 180-115=65

8 0

4 years ago

A steel safe with mass 2200 kg falls onto concrete. Just

Virty [35]

The kinetic energy of the safe increases the force exerted by the concrete

to several times the weight of the safe.

The magnitude of the force exerted on the safe by the concrete on the is approximately $\underline{29.\overline 3 \, \mathrm{MN}}$

The concrete exerts a <u>force</u> that is approximately <u>1,359.16 times the weight of the safe</u>.

Reasons:

First part

The mass of the steel safe, m = 2,200 kg

Velocity of the safe just before it hits the concrete, v = 40 m/s

The amount by which the safe was compressed, d = 0.06 m

The kinetic energy, K.E., of the safe just before it hits the round is therefore;

$\displaystyle K.E. = \mathbf{\frac{1}{2} \cdot m \cdot v^2}$

$\displaystyle K.E._{safe} = \frac{1}{2} \times 2,200 \times 40^2 = 1,760,000 \ Joules$

Work done by concrete, W = Force, F × Distance, d

$\displaystyle Force, \, F = \mathbf{\frac{Work, \, W}{Distance, \, d}}$

By the law of conservation of energy, we have;

The work done by the concrete, W = The kinetic energy, K.E. given by the safe

W = K.E. = 1,760,000 J

The effect of the work = The change in the height of the safe

Therefore;

The distance, <em>d</em>, over which the force of the concrete is exerted = The change in the height of the safe = 0.06 m

d = 0.06 m

Therefore;

$\displaystyle The \ force \ of \ the \ concrete, \, F = \frac{1,760,000\, J}{0.06 \, m} = 29,333,333. \overline 3 \, N = 29.\overline 3 \ MN$

The force of the concrete on the safe = $\underline{29.\overline 3 \ MN}$

Second part:

The gravitational force of the Earth on the safe, W = The weight of the safe

W = Mass, m × Acceleration due to gravity, g

W = 2,200 kg × 9.81 m/s² ≈ 21,582 N

The ratio of the force exerted by the concrete to the weight of the safe is found as follows;

$\displaystyle Ratio \ of \ forces = \frac{29.\overline 3 \times 10^6 \, N}{21,582 \, N} = \frac{4,000,000}{2,943} \approx \mathbf{1359.16}$

The <u>force</u> exerted by the concrete is approximately <u>1,359.16 times the weight of the safe</u>.

Learn more here:

brainly.com/question/21060171

5 0

3 years ago

If T is the centroid of ∡MNP, TN = 16, MQ = 23, and RP = 18, find each missing measure.

kramer

Answer:

a) QT = 8

b) QN = 24

c) QP = 23

d) MP = 46

e) RT = 6

f) TP = 12

Step-by-step explanation:

Find the diagram attached

If T is the centroid of<MNP, then the following are true;

TN = 2QT

Given TN = 16

16 = 2QT

QT = 16/2

QT = 8

QN = QT + TN

QN = 8 + 16

QN = 24

MQ + QP = MP and MQ = QP

If MQ = 23

QP = 23

Recall that MP = MQ + QP

MP = MQ + MQ

MP = 2MQ

MP = 2(23)

MP = 46

Similarly, TP = 2RT

RT = TP/2

Also RP = RT + TP

RP = RT + 2 RT

RP = 3RT

18 = 3RT

RT = 18/3

RT = 6

Recall that TP = 2RT

TP = 2(6)

TP = 12

3 0

3 years ago

Given h(x)=2x+5h(x)=2x+5, find h(-6)h

Fudgin [204]

Answer:

7

Step-by-step explanation:

h(6)=2(6)–5

h(6)=12–5

=7

3 0

3 years ago

Read 2 more answers

What is the solution to the equation?<br> 16 = -d + 6<br> a) 10<br> b) - 10<br> d) - 15<br> c) - 9

Rasek [7]

Answer:

Step-by-step explanation:

-d + 6 = 16

-d = 10

d = -10

the answer is b

7 0

3 years ago