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ratelena [41]
3 years ago
15

What effect does observing a substance's physical properties have on the substance?

Physics
1 answer:
Free_Kalibri [48]3 years ago
6 0

If you're careful, you ought to be able to observe ANY of these properties
without any effect on the substance:

Absorption, albedo, angular momentum, area, color, concentration,
density, elasticity, electric charge, electrical conductivity, flow rate,
electrical impedance, electric potential, fluidity, length, location, mass,
luminance, luminescence, luster, magnetic field, momentum, opacity,
permeability, permittivity, plasticity, pressure, radiance, solubility, spin,
specific heat, resistivity, reflectivity, refractive index, temperature,
thermal conductivity, velocity, viscosity, or volume.

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Are you a negro can i call you one
Roman55 [17]

Answer:

no it is mean call people their names

Explanation:

because if my name was Shar i would want people to call me Shar not negro

3 0
3 years ago
Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

7 0
3 years ago
An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo
Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

The electric field strength at point (x,y) = (0cm, 20 mm) is =<u>35321.58999 N/C</u>

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=<u>16321.0769 N/C</u>

 Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷  0.000000152881

=<u>35321.58999 N/C</u>

8 0
2 years ago
A car accelerates from rest, and travels 400 m in 3.5 seconds. If
zheka24 [161]

Answer:

A car accelerates from rest, and travels 400 m in 3.5 seconds. If

the net force on the car is 12,000 N what is the mass of the car? bzgs dvd d dv dvdvd dhd dbvd

Explanation:

shd dhd bdvd dhdbduhdbdhdbbdceudd f

3 0
3 years ago
A person pushes a large 42.9 kg box at a constant velocity of 9 m/s across a horizontal floor for 3.8 s. Find the
Gwar [14]

Answer:

10 kJ

Explanation:

W = Fd

W = (μN)(vt)

W = μ(mg)vt

W = 0.7(42.9)(9.81)(9)(3.8)

W = 10,075.12506 J

W ≈ 10 kJ

6 0
3 years ago
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