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slamgirl [31]
3 years ago
6

A 2,493-kg car is moving down a road with a slope (grade of 14% at a constant speed of 13 m/s. what is the direction and magnitu

de of the frictional force? (define positive x in the forward direction, i.e., down the slope?
Physics
1 answer:
sweet [91]3 years ago
3 0
To calculate for the frictional force, we calculate first for the normal force (forcer perpendicular to the surface)
                               Fn = (2,493 kg) (9.8 m/s²) = 24243.51 N
Then, we multiply this value with the frictional constant which is not given in this problem. If the frictional coefficient is equal to x then, the frictional force is equal to 24243.51x. 
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Each cell in the battery has a potential difference of 1,5 V. calculate. the reading on the ammeter ,A​
mr_godi [17]

Answer:

0.30A

Explanation:

5 0
3 years ago
It is summertime and the temperature is 100 degrees Fahrenheit. Johnny decides he is going to go swimming. He jumps in his pool
Aloiza [94]

Answer:

Water temperatures are slow to heat up, It takes 4 times the energy to heat up water than to heat air. Water also "feels" colder because water is a more efficent medium than air to cool our body down.

Re word it tho

Hope this helps!!

3 0
3 years ago
Identify the labeled parts in the figure.
Hitman42 [59]

Answer:

I. a, c, f and h

II. e

III. b, d, g and i

IV. i

Explanation:

I. Chemical symbols are simple abbreviations used to represent various elements or compound. They consist entire of alphabet.

For the diagram given above, the labelled parts which represent chemical symbol are: a, c, f and h

II. Coefficients are numbers written before the chemical symbol of elements or compound.

For the diagram given above, the labelled part which represent Coefficient is: e

III. Number of atoms of element present in a compound is simply obtained by taking note of the numbers written as subscript in the chemical formula of the compound.

For the diagram given above, the labelled part which represent the number of atoms of the element are: b, d, g and i

IV. When no number is written as subscript in the formula of the element in the compound, it means the element has just 1 atom in the compound.

For the diagram given above, the labelled part which indicates that only 1 atom of the element is present is: i

5 0
3 years ago
A hollow conducting sphere with an outer radius of 0.295 m and an inner radius of 0.200 m has a uniform surface charge density o
IrinaK [193]

Answer:

a. 6.032\times10^{-6}C/m^2

b.6.816\times10^5N/C

Explanation:

#Apply  surface charge density, electric field, and Gauss law to solve:

a. Surface charge density is defined as charge per area denoted as \sigma

\sigma=\frac{Q}{4\pi r_{out}^2}, and the strength of the electric field outside the sphere E=\frac{\sigma _{new}}{\epsilon _o}

Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2  #surface charge outside sphere.

\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2

Hence, the new charge density on the outside of the sphere is 6.032\times10^{-6}C/m^2

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be 6.032\times10^{-6}C/m^2,

E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C

Hence, the strength of the electric field just outside the sphere is 6.816\times10^5N/C

5 0
3 years ago
An object is removed from a room where the temperature is 69 degrees and is taken outside, where the air temperature is 30 degre
Yuliya22 [10]

Answer:

The temperature of the object at any time t, T(t) is given as

T = T∞ + (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ

Explanation:

Let T be the temperature of the object at any time

T∞ be the temperature outside = 30°

T₀ be the initial temperature of the object in the room = 69°

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the object = Rate of Heat gain by the outside air

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 30) = (69 - 30)e⁻ᵏᵗ

(T - 30) = 39 e⁻ᵏᵗ

At 1 minute, T = 52°

52 - 30 = 39 e⁻ᵏᵗ

22/39 = e⁻ᵏᵗ

- kt = In (22/39) = In (0.564)

- k(1) = - 0.5725

k = 0.5725 /min

(T - T∞) = (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ

T = T∞ + (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ

4 0
3 years ago
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