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olga nikolaevna [1]
3 years ago
6

A kid drives 4 miles to the mall. If the speed limit is 45 miles/hr and the kid makes the trip in .03 hours. Is the kid breaking

the speed limit? a. No, he is going 1 mile/hr. b. Yes, he is going 133 miles/hr, C. No, he is going 40 miles/hr. d. Yes, he is going 150 miles/hr.
Physics
1 answer:
yuradex [85]3 years ago
7 0

Answer:B

Explanation:

.03 of an hr is 2 mins and if it takes 2 mins to drive 4 miles enter it in pace calulator you are going 120mph so the closest to 120 is 133mph

You might be interested in
A thin, uniformly charged insulating rod has a linear charge density λ = 3 nC/m and lies along the x axis from x = 1m to x = 3m.
Contact [7]

Answer:

A) V_A = 11.93~V

B) The vector definition of E-field is

\vec{E} = -1.13\^x + 2.41\^y

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

V = \frac{1}{4\pi \epsilon_0}\frac{Q}{r}

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

r = \sqrt{x^2+4^2}

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

dV = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{\sqrt{x^2 + 4^2}}

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

V = \int\limits^3_1 {dV} \, dx = \frac{1}{4\pi \epsilon_0}\int\limits^3_1 {\frac{\lambda}{\sqrt{x^2 + 16}} \, dx

By using an integral table, this can be calculated:

V = \frac{3\times 10^{-9}}{4\pi\epsilon_0}\ln(|\sqrt{x^2+16}+x|)\left \{ {{x=3} \atop {x=1}} \right. \\V = 11.93~V

B) The electric field can be found by a similar approach, but a different formula:

\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\^r

Let's apply this formula to the infinitesimal part we have chosen.

dE_x = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\cos(\theta)\\dE_y = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\sin(\theta)

By the geometry sine and cosine terms can be found:

\sin(\theta) = \frac{4}{\sqrt{x^2+16}}\\\cos(\theta) = \frac{x}{\sqrt{x^2 + 16}}

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

E_x = \int\limits^3_1 {dE_x} \, dx = \frac{\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{x}{(x^2+16)^{3/2}}} \, dx  = 1.13(-\^x)\\E_y = \int\limits^3_1 {dE_y} \, dx = \frac{4\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{1}{(x^2+16)^{3/2}}} \, dx  = 2.41(\^y)

So, the final E-field is

\vec{E} = -1.13\^x + 2.41\^y

The magnitude of the E-field is

E = 2.66 N/m

6 0
3 years ago
The mass of the rock is 1220 kg. It had 400 J of potential energy when it rolled down the hill. calculate the heihht.
Sauron [17]

potential is equal to mgh

we know that mass = 1220 kg

g=9.8 0r 10

400=1220*9.8*x

400=11956x

x=400/11956

5 0
3 years ago
Read 2 more answers
A 1.20 × 104 kg railroad car moving at 7.70 m/s to the north collides with and sticks to another railroad car of the same mass t
saul85 [17]

Answer: 4.77m/s

Explanation:

According to the law of conservation of momentum which states that the sum total of momentum of bodies before collision is equal to the sum of their momentum after collision. Note that the two bodies will move at a common velocity after colliding.

Let m1 and m2 be the mass of the first and second railroad cars

u1 and u2 be the velocities of the railroad cars

v be the common velocity

Using the formula

m1u1 + m2u2 = (m1 +m2)

m1 = 1.20×10⁴kg

m2 = 1.20×10⁴kg (body of same mass)

u1 = 7.70m/s

u2 = 1.84m/s

v = ?

(1.20×10⁴×7.7) + (1.20×10⁴×1.84) = (1.20×10⁴ + 1.20× 10⁴)v

9.24×10⁴ + 2.21×10⁴ = 2.4×10⁴v

11.45×10⁴ = 2.4×10⁴v

v = 11.45×10⁴/2.4×10⁴

v = 4.77m/s

The velocity of the cars after collision will be 4.77m/s

5 0
3 years ago
The temperature of a plastic cube is monitored while the cube is pushed 3.4 m across a floor at constant speed by a horizontal f
Nesterboy [21]

Answer:

\Delta E_{floor} = 51 J

Explanation:

The work (W) done on the cube to be pushed across the floor is equal to the total thermal energy (ΔE) of the system:        

W = \Delta E_{T} = \Delta E_{cube} + \Delta E_{floor} (1)

Also, the work done on the cube by the horizontal force is giving by:

W = F \cdot d (2)  

<em>where F: force applied to the cube , d: displacement of the cube     </em>    

<em>By equaling the equations (1) and (2)</em>, we can find the thermal energy of the floor:  

\Delta E_{cube} + \Delta E_{floor} = F \cdot d

\Delta E_{floor} = F \cdot d - \Delta E_{cube}

\Delta E_{floor} = 20 N \cdot 3.4 m - 17 J  

\Delta E_{floor} = 51 J

 

So, the increase in the thermal energy of the floor is 51 J.  

Have a nice day!    

7 0
3 years ago
What meal can you get from standing on a hot sidewalk riddle answer?
kozerog [31]

the answer is <u>HOT DOG</u>

3 0
3 years ago
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