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nikitadnepr [17]
3 years ago
15

What determines an object’s velocity?

Physics
1 answer:
joja [24]3 years ago
8 0

Answer:

I think it's 3) speed and direction

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Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou
kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

  • i = current flow = 3 A
  • t = time interval for which the current flow = 4\ h = 4\times 3600\ s = 14400\ s
  • V = terminal voltage of the battery
  • R = rate of energy = 9 cents/kWh

<u>Assume:</u>

  • Q = charge transported as a result of charging
  • E = energy expended
  • C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

4 0
4 years ago
It takes a car 1.75 h to go from mile marker 115 what's the average speed of the car?
emmainna [20.7K]
The average speed of the car is

               (the mile marker he passes at the end of the 1.75 h) minus 115
divided by
               (1.75 hours) .

The unit is 'miles per hour'.
6 0
3 years ago
If the magnitude of the moment of F about line CD is 57 N·m, determine the magnitude of F.If the magnitude of the moment of F ab
bazaltina [42]

Answer:

F_ab = 260.17 N

Explanation:

Given:

- Moment of force F about CD, (M)_cd = 57 Nm

Find:

- First we will write down the position vectors of points A, B , C , D:

- We will take left and bottom most corner of cube to be the origin.

- The unit vectors i , j , k are along vertical planes and outside the plane, respectively.

- The position vectors wrt to the origin are:

                             Point A = 0.2 k

                             Point B = 0.4 i + 0.2 j

                             Point C = 0.2 j + 0.4 k

                             Point D = 0.4 i + 0.4 k

- Now we will determine the Force vector F_ab along vector AB.

                             vec (AB) = B - A = 0.4 i + 0.2 j - 0.2 k

                             unit (AB) = 0.4 i + 0.2 j - 0.2 k  / sqrt ( 0.4^2 + 2*0.2^2)

                                            = [5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

Hence,

                              vec(F_ab) = Fab*[5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

- Now, form a unit vector along the line CD:

                             vec(CD) = D - C = 0.4 i - 0.2 j

                             unit (CD) = 0.4 i - 0.2 j / sqrt ( 0.4^2 + 0.2^2)

                                           = [sqrt(5)]*(0.4 i - 0.2 j)

- Now select a point on line CD, lets say C. Find the moment arm from line of action of force along AB and line CD. Hence, vector AC is:

                               vec(AC) =r_ac = C - A = 0.2 j + 0.2 k

- Now the moment about a line CD due to force is:

                              (M)_cd = unit(CD) . ( r_ac x vec(F_ab) )

The cross product of r_ac and vec(F_ab) is as follows:

                               (M)_c =  ( r_ac x vec(F_ab) ) :

                                \left[\begin{array}{ccc}i&j&k\\0&0.2&0.2\\0.8165&0.40824&-0.40824\end{array}\right]

                              (M)_c =  F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k]

The dot product of (M)_c and unit (CD)  is as follows:

                              (M)_cd = unit(CD) . (M)_c :

 (M)_cd = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k] .  [sqrt(5)]*(0.4 i - 0.2 j)

                              (M)_cd = F_ab*(sqrt(30) / 25)

- The given magnitude of the moment is (M)_cd. Calculate F_ab:

                               57 = F_ab*(sqrt(30) / 25)  

                              F_ab = 260.17 N

7 0
3 years ago
How do the densities of the jovian and terrestrial planets compare?
timofeeve [1]
Hello

The jovian density is 1.33~g/cm^3, while the density of terrestrial planets is much higher. For instance, the earth's density is 5.51~g/cm^3. The reason of this difference is in the different chemical composition: while jupiter is made of gas (mainly hydrogen), with lower density, while the terrestrial planets are made of solid and liquid, with compounds of higher density.
5 0
3 years ago
I WILL NAME YOU BRAINLIEST IF YOU ANSWER THIS!! This question will make you think twice, let see if you can answer this! Worth 1
11Alexandr11 [23.1K]

Answer:

He took a picture (or "shot") of the buffaloes. How cute :)

7 0
3 years ago
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