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Fudgin [204]
3 years ago
9

A 58.5 kg sprinter starts a race with an acceleration of 4.40 m/s2. What is the net external force on him?

Physics
1 answer:
mote1985 [20]3 years ago
4 0

Answer:

External force on him = 257.40 N

Equation is; F = ma (where 'f' is force, 'm' is mass and 'a' is acceleration)

Explanation:

The mass of the sprinter is 58.5 kg

His acceleration is 4.40 m/s²

According to Newton's second law of motion; F = ma

External force on the sprinter = 58.5 kg × 4.40 m/s² = 257.40 N

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What is the most likely reason that Mendeleev placed tellurium before iodine?
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Mendeleev watched that tellurium has compound properties like different components in its gathering, and he didn't realize that neutrons cause the more noteworthy nuclear mass. Mendeleev expressed that he anticipated that tellurium would have a lower nuclear mass than iodine.
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A ball is thrown straight up with a speed of 30
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A ball is thrown straight up with a speed of 30

m/s. What is the maximum height reached by

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During the first stage of a multistage rocket, a satellite is:
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B. sent through the atmosphere 

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The volume of an ideal gas is adiabatically reduced from 151 L to 80.6 L. The initial pressure and temperature are 1.50 atm and
Zolol [24]

Answer:

gas is dioatomic

T_f = 330.0 K

\eta = 7.07 mole

Explanation:

Part 1

below equation is used to determine the type Gas by determining \gamma value

\frac{V_{1}}{V_{F}}\gamma=\frac{P_{i}}{P_{f}}

where V_i and V_f is initial and final volume respectively

and P_i and P_f are initial and final pressure

\gamma = \frac{ln(P_f/P_i)}{ln(V_i/V_f)}

\gamma = \frac{ln(3.61/1.50)}{ln(151/80.6}

\gamma = 1.38

therefore gas is dioatomic

Part 2

final temperature in adiabatic process is given as

T_f = T_i*[\frac{v_i}{V_f}](^\gamma-1)

substituing value to get final temperature

T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}

T_f = 330.0 K

Part 3

determine number of moles by using following formula

\eta =\frac{PV}{RT}

\eta =\frac{1.013*10^{5}*0.151}{8.314*260}

\eta = 7.07 mole

4 0
3 years ago
A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition
Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So mg=Kx

K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m

Now we know that \omega =\sqrt{\frac{K}{m}}

So \frac{2\pi }{T} =\sqrt{\frac{K}{m}}

\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}

T=2.1371sec

8 0
3 years ago
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