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2 years ago

An 88.0 kg sprinter starts a race with an acceleration of 1.60 m/s2. What is the net external force (in N) on him

Physics

1 answer:

aleksandrvk [35]2 years ago

3 0

Answer:

The correct answer is "140.8 N".

Explanation:

Given:

Mass,

m = 88 kg

Acceleration,

a = 1.60 m/s²

Now,

The net external force for him will be:

⇒ $F_{ext}=Mass\times Acceleration$

By putting the values, we get

$=88\times 1.60$

$=140.8 \ N$

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Sidana [21]

Answer:

T = 273 + (-50) = 273 – 50 = 223 K

R = 188.82 J / kg K for CO2

Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg / $m^{3}$

T = 273 +18 = 291 K, R = 287 J / kg k (for air) P = 101.6 k Pa = 101600 Pa

Density (Earth Atmosphere) = P / RT = 101600 / 287 x 291 = 1.216 kg / $m^{3}$

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2 years ago

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A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f

Snezhnost [94]

Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

$\mu mg = m\frac{v^2}{r}$

where

$\mu$ is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

$v=\sqrt{\mu gr}$

And by substituting the data of the problem, we find the speed at which the car begins to skid:

$v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s$

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2 years ago

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The weight of Earth's atmosphere exerts an average pressure of 1.01 ✕ 105 Pa on the ground at sea level. Use the definition of p

zloy xaker [14]

Answer:

The weight of Earth's atmosphere exert is $516.6\times10^{17}\ N$

Explanation:

Given that,

Average pressure $P=1.01\times10^{5}\ Pa$

Radius of earth $R_{E}=6.38\times10^{6}\ m$

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

$P=\dfrac{F}{A}$

$F=PA$

$F=P\times 4\pi\times R_{E}^2$

Where, P = pressure

A = area

Put the value into the formula

$F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2$

$F=516.6\times10^{17}\ N$

Hence, The weight of Earth's atmosphere exert is $516.6\times10^{17}\ N$

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2 years ago

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A tuning fork produces a sound with a frequency of 256 hz and a wavelength in air of 1.33 m. find the speed of sound in the vici

aalyn [17]

<span>We can think this through intuitively. A frequency of 256 Hz means that the wave has 256 cycles each second. If the wavelength is 1.33 meters, then there are 256 of them each second. Therefore, we just need to multiply the wavelength by the frequency to find the speed of sound. (Note that the units Hz = 1 / s) v = (frequency) x (wavelength) v = (256 Hz) x (1.33 m) v = 340.5 m/s The speed of sound in the vicinity of the fork is 340.5 m/s</span>

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2 years ago

A na+ ion moves from inside a cell, where the electric potential is -72 mv, to outside the cell, where the potential is 0 v. wha

Vlada [557]

The change in electric potential energy of the ion is equal to the charge multiplied by the voltage difference:
$\Delta U = q \Delta V = q (V_f - V_i)$
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Substituting the numbers, we find:
$\Delta U = (1.6 \cdot 10^{-19}C)(0 V-(-0.072 V))=1.15 \cdot 10^{-20} J$

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2 years ago

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