If you know that a person is running 100 feet every 12 seconds, you can determine their __________.

1 answer:

6 0

<span>If you know that a person is running 100 feet every 1 seconds, you can determine their velocity. Velocity is the rate of motion. i.e. the rate at which the person is running. Velocity is given by distance / time. In the given question the distance is 100 feet while the time is 12 seconds. So to determine the velocity, we divide 100 by 12 to get 8.33 feet per second.</span>

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Mrs. Waid drives her car up Route 8 at 80mph. If her car weighs 2500 lbs,

Anvisha [2.4K]

1.358 J is the kinetic energy of the car driven by Mrs. Waid.

<u>Explanation:</u>

Given data:

Velocity at which Mrs. Waid drives her car = 80 mph

In order to convert mph (meter per hour) into mps (meter per second),

$\frac{80}{3600}=0.0222 \mathrm{m} / \mathrm{s}$

Car weighs 2500 lbs, means mass of the car, m = 2500 lbs

I kilo gram = 2.20462 pound

Therefore, 1 pound (lb)= 0.45359237 kilograms (kg).

To converting pounds into kilogram,

$\frac{2500}{0.45359237}=5511.55 \mathrm{kg}$

As we know, the kinetic energy can be defined as directly proportionate to the object’s mass (m) and square of its velocity (v). The expression can be given as below,

$\text { kinetic energy }(K . E)=\frac{1}{2} \times m \times v^{2}$

By substituting the given values, we get

$\text { kinetic energy }(K . E)=\frac{1}{2} \times 5511.55 \times 0.0222^{2}$

$\text { kinetic energy }(K . E)=\frac{1}{2} \times 5511.55 \times 0.00049$

$kinetic energy (K . E)=1.358 joule$

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4 years ago

A taxi

nignag [31]

That taxi will traveled 1500s by carrying the passenger.

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3 years ago

A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2

Leno4ka [110]

Answer:

A. T=15.54 °C

B. Q/A= 0.119 W/m2

Explanation:

To solve this problem we need to use the Fourier's law for thermal conduction:

$Q= kA\frac{dT}{dx}$

Here, the rate of flow per square meter must be the same through the complete wall. Therefore, we can use it to find the temperature at the plane where the wood meets the Styrofoam as follows:

$\frac{Q}{A} =\frac{T_1-T_0}{d_w}k_w=\frac{T_2-T_1}{d_s}k_s\\T_1(\frac{k_w}{d_w}+\frac{k_s}{d_s})=T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}\\T_1=\frac{T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}}{\frac{k_w}{d_w}+\frac{k_s}{d_s}}\\T_1= 15.54 \°C$

Then, to find the rate of heat flow per square meter, we have:

$\frac{Q}{A}=\frac{T_1-T_0}{d_w}k_w=0.119 \frac{W}{m^2}\\\frac{Q}{A}=\frac{T_2-T_1}{d_s}k_s= 0.119 \frac{W}{m^2}$

$T_0: Temperature \ in \ the \ house\\T_1: Temperature \ at \ the \ plane \ between \ wood \ and \ styrofoam\\T_2: Temperature \ outside\\k_w: k \ for \ wood\\d_w: wood \ thickness\\k_s: k \ for \ styrofoam\\d_s: styrofoam \ thickness$