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OlgaM077
3 months ago
15

A car covers 72 kilometers in the first hour of its journey. In the next hour, it covers 90 kilometers. What is the amount of wo

rk done by the car? The total mass of the car, including its passengers, is 2.5 × 103 kilograms.
A. 2.8 × 105 joules
B. 4.0 × 105 joules
C. 5.1 × 105 joules
D. 2.8 × 106 joules
E. 2.8 × 107 joules
Physics
2 answers:
Sav [3.1K]3 months ago
4 0

Answer:

(1/2)*2500*(25^2) - (1/2)*2500*(20^2) = 281250 J, which is also equivalent to 2.8 x 10^5 Joules.

Explanation:

Softa [3K]3 months ago
4 0
The work done can be calculated using the equation:

Work = Force x Distance = Change in kinetic energy

The kinetic energy is derived using the following formula: KE = (1/2)*m*v^2
Thus, the change in kinetic energy is calculated as (1/2)*m*(Vf)^2 - (1/2)*m*(Vo)^2

Where:

Vf represents the final speed = 90 kph = 25 m/s
Vo denotes the initial speed = 72 kph = 20 m/s

By substituting in the given values:

Work = (1/2)*2500*(25^2) - (1/2)*2500*(20^2) = 281250 J, which can also be represented as 2.8 x 10^5 Joules.

The correct choice among the options is A.
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You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a
serg [3582]

Answer:

The accurate statements are

2. The train is not an inertial frame of reference.

5. The train could be moving at a constant velocity in a circular path.

8. The train must be undergoing acceleration.

Explanation:

As we observe that the string forms an angle with the horizontal

we can formulate the force equation relevant to the given ball

F_x = Tcos\theta

ma = Tcos45

similarly in the Y direction

mg = Tsin45

Thus we conclude

\frac{ma}{mg} = cot 45

a = g cot45

This leads us to deduce that the train is accelerating with an acceleration identical to that of gravity

The correct statements will be

2. The train is not an inertial frame of reference.

5. The train could be moving at a constant speed in a circular path.

8. The train must be experiencing acceleration.

8 0
3 months ago
Consider the uniform electric field E = (2.5 j + 3.5 k) × 103 N/C. (a) Calculate the electric flux through a circular area of ra
kicyunya [3294]
The desired electric flux Φ is calculated from the electric field E=(2.5• j + 3.5• k) ×10³ N/C and a circular path with a radius r=2.5m. The electric flux across a surface is represented as Φ=∮E•dA. Given the area A lies in the yz-plane, the normal orientation flows in the x-direction with A=πr² leading to dA=2πrdr •i. Thus, Φ evolves to Φ=∮(2.5j + 3.5k)×10³•(2πrdr i). Integrating from r=0 to r=2.5m and noting that components in different directions yield zero, results in Φ equaling 0 Nm²/C. Regarding the second part, when the area vector is at a 45° angle to the xy-plane, we redefine dA as (2πrCos45 i + 2πrSin45 j) dr, leading to the new flux calculation as Φ=10³∮ 5πrSin45 dr, integrating from 0 to 2.5m. With substitutions made, the result comes to Φ=34.71×10³ Nm²/C.
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3 months ago
The filament of the light bulb is made of tungsten. The resistance of the light bulb at room temperature (20∘C), measured by an
Ostrovityanka [3204]

Explanation:

The formula illustrating the relationship between resistance and temperature is as follows:

R =

R_{o} + \alpha [T_{2} - T_{1}]

where,   R = final resistance

       

= initial resistanceR_{o}

       

= temperature coefficient of resistivity\alpha

       

= final temperature     T_{2}

       

= initial temperatureT_{1}

Given data as follows.

     

T_{1} = (20 + 273) K = 293 K      

      R = 36 ohm,  

= 3 ohmT_{2}

         

= 0.0045R_{o}

Substituting the provided values into the above formula gives us the following.

        R = \alpha

        36 =

R_{o} + \alpha [T_{2} - T_{1}]      

=

3 + 0.0045 \times [T_{2} - 293]

                 = 7626.33 K

T_{2}Thus, it can be concluded that \frac{34.3185}{0.0045}the temperature of the light bulb at 12.0 V is 7626.33 K.

7 0
3 months ago
According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f
ValentinkaMS [3465]

Answer:

S_{s}=300 m/s

According to the guideline for kilometers, every three seconds between a lightning strike and the subsequent thunder indicates the distance to the flash in kilometers.

Explanation:

To calculate the speed of sound in meters per second, we need to utilize certain conversion factors. One mile corresponds to 5 seconds after witnessing the lightning. Furthermore, 1 mile comprises 5280 feet, and 1 foot is equivalent to 0.3048 meters. This information is sufficient to solve the issue. The conversion ratios can be set up like this:

\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s

Observe how the ratios are organized such that the units cancel out during calculations. One ratio has miles in the numerator while the other has them in the denominator, leading to cancellation. The same applies to the feet.

The question requires us to provide the answer to one significant figure, resulting in the speed of sound rounding to 300m/s.

For the second part, we will again utilize conversions. This time we will set our ratios in reverse and realize that there are 1000 meters in 1 kilometer, leading us to:

\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km

This signifies that for every 3.11 seconds, the distance to the lightning strike is 1 kilometer. Since this is a fabric of general knowledge, we round to the nearest whole number for simplicity, establishing the guideline:

According to the rule for kilometers, every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

3 0
3 months ago
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