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igor_vitrenko [27]

2 years ago

14

Ben walks 500 meters from his house to the corner store. He then walks back toward his house, but continues 200 meters past his

house to talk to a neighbor. It takes Ben 17 minutes from the time he leaves his house until he stops to talk to his neighbor. What is Ben’s average velocity?

2 answers:

Oksana_A [137]2 years ago

6 0

71 MPM (Meters Per Minute)
S = Speed
D = Distance
T = Time
to find the Speed you divide D by your T

liq [111]2 years ago

6 0

Answer:

0.196 m/s

Explanation:

Average displacement is defined as the total displacement covered to the total time taken.

Total displacement = 500 - 500 - 200 = - 200 m

Total time taken = 17 minutes = 17 x 60 = 1020 second

So, Average velocity = 200 / 1020 = 0.196 m/s (magnitude only)

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A vertical spring of constant k = 400 N/m hangs at rest. When a 2 kg mass is attached to it, and it is released, the spring exte

Viefleur [7K]

Answer:

4.9 cm

Explanation:

From Hook's Law,

F = ke......................... Equation 1

Where F= force, e = extension, k = spring constant.

Note: the Force acting on the the spring is the weight of the mass.

W = mg.

F = mg.................... Equation 2

Where m = mass, g = acceleration due to gravity

Substitute equation 2 into equation 1

mg = ke

make e the subject of the equation

e = mg/k............... Equation 3.

Given: m = 2 kg, g = 9.8 m/s², k = 400 N/m

e = (2×9.8)/400

e = 19.6/400

e = 0.049 m

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3 0

2 years ago

In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into t

Pie

Answer:

A) $Q_a=74256\ J$

B) $Q=93562560\ J$

Explanation:

Given:

temperature of air, $T_a=-19+273=254\ K$

temperature of lungs, $T_l=37+273=310\ K$

specific Heat exchanged from the lungs , $c_l=0.47\ J.kg^{-1}.K^{-1}$

specific heat of air, $c_a=1020\ J.kg^{-1}.K^{-1}$

mass of 1 L air, $m'=1.3\ kg$

breath rate, $b=21\ breath.min^{-1}$

A)

Now,

amount of heat needed to warm the air of lungs to the body temperature:

$Q_a=m'.c_a.\Delta T$

$Q_a=1.3\times1020\times (310-254)$

$Q_a=74256\ J$

B)

Amount of heat lost per hour:

<u>No. of breaths per hour:</u>

$B=b.60$

$B=21\times 60$

$B=1260$

<u>Now the total loss of energy in 1 hr.:</u>

$Q=Q_a.B$

$Q=74256\times 1260$

$Q=93562560\ J$

7 0

2 years ago

In 1923, the United States Army (there was no U.S. Air Force at that time) set a record for in-flight refueling of airplanes. Us

dybincka [34]

Answer:

1.95m/s

Explanation:

Please view the attached file for the detailed solution.

The following were the conversion factors used in order to express all quatities in SI units:

$1 gallon=0.00378541m^3\\1 inch=0.0254m\\1 minute=60s$

6 0

2 years ago

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes

OlgaM077 [116]

Answer:

Explanation:

mass of car, m = 1000 kg

initial velocity, u = 20 m/s

final velocity, v = 0 m/s

distance, s = 120 m

Let a be the acceleration of motion

use third equation of motion

v² = u² + 2 as

0 = 20 x 20 + 2 x a x 120

a = - 1.67 m/s²

Let F be the force

Force, F mass x acceleration

F = - 1000 x 1.67

F = - 1666.67 N

The direction of force is towards south and the magnitude of force is 1666.67 N.

8 0

2 years ago

If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it

Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

$\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

Explanation:

From the question we are told that

The mass of the rocket is m = 6300 kg

The velocity at gas is being ejected is u = 2000 m/s

The initial acceleration desired is $a = 27.0 \ m/s$

The time taken for the gas to be ejected is t = 1 s

Generally this desired acceleration is mathematically represented as

$a = \frac{u * \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}$

Here $\frac{\Delta m}{\Delta t }$ is the rate at which gas is being ejected with respect to time

Substituting values

$27 = \frac{2000 * \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}$

=> $170100 -27* \frac{\Delta m}{\Delta t} = 2000 * \frac{\Delta m}{\Delta t}$

=> $170100 = 2027 * \frac{\Delta m}{\Delta t}$

=> $\frac{\Delta m}{\Delta t} = \frac{170100}{2027}$

=> $\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

3 0

2 years ago