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2 years ago

11

Holding health care personal to lesser standards of care in emergencies is call the what statue?

1 answer:

olga55 [171]2 years ago

3 0

Hello there,

Holding health care personal to lesser standards of care in emergencies is call the what statue?

Answer: A. Good Samaritan

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Of these nonmetals, which one is likely to be the least reactive?

umka21 [38]

Answer:

helium

Explanation:

8 0

3 years ago

Read 2 more answers

Technician A says some compressor service procedures can be performed on the vehicle if space permits. Technician B says modern

stiv31 [10]

Answer:

Technicians A.

Explanation:

Since air compressor uses series of processes that turn incoming ambient air into a power source for tools and machinery. This means that air compressor has many different parts, and each of these parts must be maintained to ensure they function properly and optimally.

These are the basis when it comes to servicing a compressor

You need to change its oil

And clean its filters.

Inspected it's filters every three months, and have its filters replaced and connections tightened at least once every year.

To do all these can be performed on the vehicle if there is enough space just as Technician A said for the question context.

5 0

3 years ago

Suppose you have a Frisbee with a mass of 0.10 kg. You first throw it with a force of 5N and then change the force you throw it

natali 33 [55]

Answer:

The acceleration increases.

Explanation:

From Newton's 2nd Law, we have $\Sigma F=ma$ . We can see that force is directly proportional to mass and acceleration. Therefore, as force increases, either mass or acceleration must increase as well, and vice versa. Since mass is maintained here, if you increase the force applied to the Frisbee, the acceleration will increase as well.

8 0

3 years ago

A proton moves with a velocity of v with arrow = (4î − 6ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î

nalin [4]

Answer:

F = [(6.4 × 10⁻¹⁹)î + (8.0 × 10⁻¹⁹)ĵ + (22.4 × 10⁻¹⁹)k] N

Magnitude of F = (2.466 × 10⁻¹⁸) N

Explanation:

The magnetic force, F, on a given charge, q, moving with velocity, v, in a magnetic field, B, is given as the vector product

F = qv × B

where v = (4î − 6ĵ + k) m/s

B = (î + 2ĵ − k) T

The particle is a proton, hence,

q = (1.602 × 10⁻¹⁹) C

F = qv × B = q (v × B)

(v × B) is given as (4î − 6ĵ + k) × (î + 2ĵ − k)

The cross product is evaluated as a determinant of

| î ĵ k |

|4 -6 1 |

|1 2 -1 |

î [(-6)(-1) - (2)(1)] - ĵ [(4)(-1) - (1)(1)] + k [(4)(2) - (-6)(1)]

î (6 - 2) - ĵ (-4 - 1) + k (8 + 6) = (4î + 5ĵ + 14k)

(v × B) = (4î + 5ĵ + 14k)

F = q (v × B) = (1.6 × 10⁻¹⁹) (4î + 5ĵ + 14k)

F = [(6.408 × 10⁻¹⁹)î + (8.01 × 10⁻¹⁹)ĵ + (22.428 × 10⁻¹⁹)k] N

Magnitude of F =

√[(6.408 × 10⁻¹⁹)² + (8.01 × 10⁻¹⁹)² + (22.428 × 10⁻¹⁹)²]

Magnitude of F = (2.466 × 10⁻¹⁸) N

Hope this Helps!!!

4 0

2 years ago

Read 2 more answers

What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?

sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

$F=k\dfrac{q_1q_2}{d^2}$

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

$F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F$

So, the new force becomes 4 times the initial force.

4 0

2 years ago