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Sindrei
5 days ago
7

During a workout at the gym, you set the treadmill at a pase of 55.0 m/min. How many minutes will you walk if you cover a distan

ce of 7500 ft? ...?
Physics
2 answers:
Yuliya22 [2.9K]5 days ago
7 0
t = 41.56 min. To clarify, the distance covered on the treadmill is found using... Knowing that the distance and speed are involved here, we reach the conclusion. Therefore, computing the time needed to traverse the specified distance results in...
inna [2.7K]5 days ago
5 0
Here’s the solution for the posed question. If your treadmill speed is set to 55 meters per minute and you need to determine the time it takes to walk 7500 feet, you first convert 7500 feet into meters, resulting in 2286 meters. Dividing this distance by the speed of 55 meters per minute gives us 41.56. Therefore, the total time required to walk 7500 feet is approximately 41.6 minutes. I trust this information is helpful.
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The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes
Sav [2826]

mass₃<mass₁=mass₅<mass₂=mass₄

Explanation:

Data points:-

1. mass:  m      speed: v

2. mass: 4 m   speed: v

3. mass: 2 m   speed: ¼ v

4. mass: 4 m   speed: v

5. mass: 4 m   speed: ½ v

We know that the formula for Kinetic energy (KE) is ½ mv²

Where m represents the mass of the object

           v represents the object's velocity

<psubstituting the="" given="" values="" for="" mass="" and="" speed="" from="" previous="" data:="">

The KE of Body 1(mass₁) = ½*m*v²             = mv²/2

KE of Body 2(mass₂) = ½*4m*v²         = 2mv²

KE of Body 3(mass₃) = ½*2m*(1/4v)²  =  mv²/16

KE of Body 4(mass₄) = ½*4m*v ²        =  2mv ²

KE of Body 5(mass₅) = ½*4m*(1/2v)²  =   mv²/2

</psubstituting>
6 0
27 days ago
Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Ostrovityanka [2814]
This is somewhat misleading, and I encountered the same question in my homework. An electric field strength of 1*10^5 N/C is provided, along with a drag force of 7.25*10^-11 N, and the critical detail is that it maintains a constant velocity, indicating that the particle is in equilibrium and not accelerating.
<span>To solve, utilize F=(K*Q1*Q2)/r^2 </span>
<span>You'll want to equate F with the drag force, where the electric field strength translates to (K*Q2)/r^2; substituting the values results in </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
3 0
15 days ago
Read 2 more answers
A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation
Ostrovityanka [2814]

Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given the data:

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The heat generation formula can be articulated as follows:

q = πr²Lq'

q = π. 0.1². L. 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Applying the energy balance equation,

Energy In = Energy Out

This equates to q, which is 754L

From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C

Additionally, the outer surface temperature records as 51° C

Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C

6 0
1 month ago
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
ValentinkaMS [3084]
No one is going to handle that for a mere 5 points lol.
8 0
1 month ago
Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?
Ostrovityanka [2814]

Answer:

2.32\times 10^{-11}

Explanation:

The first number is 4.48\times 10^{-8}.

The second number is 5.2\times 10^{-4}.

We must multiply these two numbers together.

4.48\times 10^{-8}\times 5.2\times 10^{-4}=(4.48\times 5.2)\times 10^{(-8-4)}\\\\=23.296\times 10^{-12}

In scientific notation: 2.32\times 10^{-11}

Therefore, this is the solution you are looking for.

8 0
1 month ago
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