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Soloha48 [4]
4 years ago
12

You would have the largest mass of gold if your chunk of gold weighed 1 N on

Physics
2 answers:
Neporo4naja [7]4 years ago
6 0
The answer is 1 pound

SSSSS [86.1K]4 years ago
5 0

The answer is Jupiter.

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In the diagnostic radiology energy range (which includes mammography) from 23 to 150 kVp, which of the following tissues possess
Semmy [17]

Answer:

Bone

Explanation:

Diagnostic radiology include the use of non-invasive imaging scans to diagnose a patient.

The voltages used in diagnostic tubes range from roughly 20 kV to 150 kV and thus the highest energies of the X-ray photons range from roughly 20 keV to 150 keV.

The tests and equipment used sometimes involves low doses of radiation to create highly detailed images of an area.

4 0
3 years ago
A force of 10N is required to stretch a spring from 20cm to 25cm. What is the spring constant in N/m2 Be careful of unit
kupik [55]

Answer:

C) 40 N/m

Explanation:

If we ASSUME that the spring is un-stretched at the zero cm position

k = F/Δx = 10/0.25 = 40 N/m

5 0
3 years ago
Streams pick up a greater load of sand and gravel as they leave a mountain range and flow across a lowland.
inessss [21]
B

because it will pick it up while coming down not just flowing on low land<span />
6 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
If a reaction is exothermic, then which of the following statements is always true? If a reaction is exothermic, then which of t
sesenic [268]

Explanation:

Reactants ⇒ product + heat (exothermic reaction) ....(1)

We have given that the reaction is exothermic, so the heat is rejected from the reaction. We know that heat is the form of energy.

From equation (1)

from the given equation we can see that reactants have more energy than products.

So the reactants have higher potential energy in comparison to the products.

7 0
3 years ago
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