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klasskru [66]
3 years ago
8

suppose you are walking home after school. The distance from school to your home is five kilometers. on foot you can get home in

25 minutes. However if you rode a bicycle you could get gome in 10 minutes. what is your average speed while walking
Physics
2 answers:
SSSSS [86.1K]3 years ago
8 0

<u>Answer:</u> The average speed while walking is 12 km/hr.

<u>Explanation:</u>

Average speed is defined as the ratio of total distance traveled to the total time taken.

To calculate the average speed while walking, we use the equation:

\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}

We are given:

Distance from school to home = 5 km

Time taken = 25 mins  = 0.417 hr   (Conversion factor: 1 min = 60 sec)

Putting values in above equation, we get:

\text{Average speed while walking}=\frac{5km}{0.417hr}=12km/hr

Hence, the average speed while walking is 12 km/hr.

agasfer [191]3 years ago
4 0
If it takes ten minutes to travel 5 km, then it'll take 60 minutes to travel thirty, making your biking speed 30 kilometers per hour. 25 minutes divided by ten is 2.5. multiply 5 by 2.5 to get 12.5, your average walking speed.
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A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st
Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r

where,

\tau= shear stress at a distance 'r' from the center

T = is the applied torque

I_{p} = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

\tau _{max}=\frac{4}{3}\times \frac{V}{A}

Applying values we get

\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa

3 0
3 years ago
A weightlifter curls a 30 kg bar, raising it each time a distance of 0.60 m. How many times must he repeat this exercise to burn
Sholpan [36]

Answe

given,

mass of the bar, m = 30 Kg

distance of rise, h = 0.60 m

Assuming the efficiency = 25 %

energy from the pizza slice = 300 C = 1260 kJ

To consume Energy from the pizza bar is to be pulled several number of time.

( energy from pizza ) x (efficiency) = n m g h

n is the number of lift

( 1260 x 10³) x (0.25) = n x 30 x 9.8 x 0.6

n = \dfrac{315\times 10^3}{176.4}

     n = 1786 times.

Weightlifter should lift bar 1786 times to burn off the energy.

8 0
3 years ago
A model engine accelerated forward from rest along a straight track at 10.0 m/s2 for 3.0 seconds. It then accelerates coward at
Mrac [35]

Answer:

17.2 seconds

Explanation:

Given:

v₀ = 0 m/s

a₁ = 10.0 m/s²

t₁ = 3.0 s

a₂ = 16 m/s²

t₂ = 5.0 s

a₃ = -12 m/s²

v₃ = 0 m/s

Find: t

First, find v₁:

v₁ = a₁t₁ + v₀

v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)

v₁ = 30 m/s

Next, find v₂:

v₂ = a₂t₂ + v₁

v₂ = (16 m/s²) (5.0 s) + (30 m/s)

v₂ = 110 m/s

Finally, find t₃:

v₃ = a₃t₃ + v₂

(0 m/s) = (-12 m/s²) t₃ + (110 m/s)

t₃ = 9.2 s

The total time is:

t = t₁ + t₂ + t₃

t = 3.0 s + 5.0 s + 9.2 s

t = 17.2 s

Round as needed.

4 0
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erastova [34]

Answer:

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Explanation:

Flutter is a type of arrhythmia that causes very fast and regular ryth of the atria of about 250 beats per minute.

Arrhythmia can be defined as any sort of irregularity heart rate or rhythm is also called as dysrhythmia.

Arrhythmias can be categorized as heart block, bradycardia, tachycardia, fibrillation, flutter, sick sinus syndrome, and is diagnosed by Electrocardiography.

In Flutter, the heart chambers do get sufficient time to get filled with blood completely prior to next contraction.

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