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3 years ago

12

Which source is most likely an authority on the subject? Dr. Vale, a pediatrician, discusses electrical circuitry on her home-im

provement website. Jonah, an amateur cook, teaches his son how to make the best grilled cheese sandwiches. Susan, an architect, presents her hand-made quilts at the Quilting Society. Dr. Patel, a cardiologist, presents her findings on heart disease.

2 answers:

Pavlova-9 [17]3 years ago

7 0

Answer:

D.

Explanation:

I took the quiz

Naddika [18.5K]3 years ago

6 0

The correct answer is Dr. Patel, a cardiologist, presents her findings on heart disease.

A cardiologist is a heart doctor, so he would be the highest authority to speak about heart disease. The others are all amateurs in their businesses.

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Answer:

1.A compound is when two substances are put together and there is a reaction creating a new substance Ex. H20. A mixture is when two substances are put together, there is no reaction and they can be separated easier than compounds. Ex. salad dressing.

2.Atomic bonds

3.Compound

4.Element

5.Mixture

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2 years ago

Pourquoi doit on toujours préciser le référentiel dans le quel est étudié le mouvement d'un système?

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The movement of an object depends on the reference frame, so it is important to predicate it.

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2 years ago

800 joules of work were done with a force of 200 newtons. Over what

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Answer:

4m

Explanation:

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3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m

nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

$F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha$

Where;

$F_T$ = The tangential force

I = The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

$I_m$ = The rotational inertia of the merry-go-round

$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Which gives;

$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

$v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5$

The velocity of the merry-go-round, $v_m$ , after the boy hops on the merry-go-round = 1.5 m/s.

5 0

2 years ago

ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit

horrorfan [7]

Answer

given,

D = 50 mm = 0.05 m

d = 10 mm = 0.01 m

Force to compress the spring

$F = \dfrac{d^4G\delta}{8D^3N}$

$\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm$

$F = \dfrac{d^4G}{8D^3}\times 0.004$

$F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004$

F = 3160 N

stress correction factor from stress correction curve is equal to 1.1

now, calculation of corrected stress

$\tau = \dfrac{8FDk_s}{\pi d^3}$

$\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}$

= 442.6 Mpa

The tensile strength of the steel material of ASTM A229 is equal to 1300 Mpa

now,

$\tau_s \leq 0.45 S_u$

$\tau_s \leq 0.45 \times 1300$

$\tau_s \leq 585\ Mpa$

since corrected stress is less than the $\tau_s$

hence, spring will return to its original shape.

6 0

3 years ago

Read 2 more answers