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wariber [46]
3 years ago
5

What is the electric field strength 7.50 x 10E-1 meters from a 2.8 x 10 E-3 C charged object?

Physics
1 answer:
Nadusha1986 [10]3 years ago
5 0

Answer:hi

Explanation:

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I NEED HELP
Oduvanchick [21]

Answer:

the answer is different

Explanation:

i took the test

3 0
3 years ago
. The artificial sweetener NutraSweet is a chemical called aspartame (C14H18N2O5). What is (a) its molecular mass (in atomic mas
loris [4]

The molecular mass, in atomic mass unit, of aspartame would be 294 amu while the mass, in kg, of an aspartame molecule would be 0.294 kg

Aspartame has the chemical formula C14H18N2O5

C = 12, H = 1, N = 14, O = 16

(a) Molecular weight = (12x14) + (1x18) + (14x2) + (16x5)

                                  = 168 + 18 + 28 + 80

                                    = 294 amu

(b) Mass of 1 molecule of aspartame = mole x molar mass

                                    = 1 x 294

                                        = 294 g

Converting 294 g to kg = 294/1000

                                         = 0.294 kg

More on mole and molar mass can be found here: brainly.com/question/6613610

7 0
2 years ago
(4. A bus is moving at 25 m/s when the driver steps on the brakes and brings the bus to a stop in
zlopas [31]

Answer:

(a)  the average acceleration of the bus while braking is 8.333 m/s

(b) if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a.  [¹/₂ (8.333 m/s) = 4.16 s]

Explanation:

Given;

initial velocity of the bus, v = 25 m/s

time of the motion, t = 3 s

(a)  the average acceleration of the bus while braking

a = dv/dt

where;

a is the bus acceleration

dv is change in velocity

dt is change in time

a = 25 / 3

a = 8.333 m/s

(b) If the bus took twice as long to stop, the duration = 2 x 3s

a = 25 / (2 x 3s)

a = ¹/₂ x (25 / 3)

a = ¹/₂ (8.333 m/s) = 4.16 s

Thus, if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a.

6 0
3 years ago
Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of
Radda [10]

According to the information provided to define an average density, it is necessary to use the concepts related to mass calculation based on gravitational constants and radius, as well as the calculation of the volume of a sphere.

By definition we know that the mass of a body in this case of the earth is given as a function of

M = \frac{gr^2}{G}

Where,

g= gravitational acceleration

G = Universal gravitational constant

r = radius (earth at this case)

All of this values we have,

g = 9.8m/s^2\\G  = 6.67*10^{-11} m^3/kg*s^2\\r = 6378*10^3 m

Replacing at this equation we have that

M = \frac{gr^2}{G} \\M = \frac{(9.8)(6378*10^3)^2}{6.67*10^{-11}} \\M = 5.972*10^{24}kg

The Volume of a Sphere is equal to

V = \frac{4}{3}\pi r^3\\V = \frac{4}{3} \pi (6378*10^3)^3\\V = 1.08*10^{21}m^3

Therefore using the relation between mass, volume and density we have that

\rho = \frac{m}{V}\\\rho = \frac{5.972*10^{24}}{1.08*10^{21}}\\\rho = 5.52*10^3kg/m^3

6 0
3 years ago
Which of these objects would be most likely to create this sound wave?
babunello [35]

Answer: It’s C.

Explanation:

4 0
3 years ago
Read 2 more answers
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