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Yanka [14]
3 years ago
12

To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp

ecified tile surface. A typical athletic shoe has a coefficient of 0.830. In an emergency, what is the minimum time interval in which a person starting from rest can move 3.20 m on a tile surface if she is wearing the following footwear?
Physics
1 answer:
motikmotik3 years ago
8 0

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that f_{max}=\mu N

Now, we find the acceleration:

\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s

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xz_007 [3.2K]
Thermal energy is transforming, i think.
5 0
3 years ago
An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th
neonofarm [45]

Answer:

Electric force, F=2.24\times 10^{-14}\ N

Explanation:

It is given that,

Charge on an electron is -1.6\times 10^{-19}\ C

Electric field, E=1.4\times 10^5\ N/m

We need to find the magnitude of the electric force on this electron due to this field. The electric force is given by :

F=qE\\\\F=1.6\times 10^{-19}\times 1.4\times 10^5\\\\F=2.24\times 10^{-14}\ N

So, the electric force is 2.24\times 10^{-14}\ N.

6 0
2 years ago
Joan stirred a pot of hot soup with a metal spoon. after a minute or two, the handle of the spoon became too hot to hold. what k
natima [27]

When the metal spoon comes in contact with the hot soup, heat transfer from hot soup to the spoon because of difference in temperature of the two. The kind of heat transfer represented here is conduction. In conduction, heat is transferred from a region of higher temperature to a region of lower temperature when there is a physical contact between the two.

4 0
3 years ago
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on
disa [49]

Answer:

V = 48 Volts

Explanation:

Since we know that electric potential is a scalar quantity

So here total potential of a point is sum of potential due to each charge

It is given as

V = V_1 + V_2 + V_3

here we have potential due to 50 nC placed at y = 6 m

V_1 = \frac{kQ}{r}

V_1 = \frac{(9\times 10^9)(50 \times 10^{-9})}{\sqrt{6^2 + 8^2}}

V_1 = 45 Volts

Now potential due to -80 nC charge placed at x = -4

V_2 = \frac{kQ}{r}

V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}

V_2 = -60 Volts

Now potential due to 70 nC placed at y = -6 m

V_3 = \frac{kQ}{r}

V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}

V_3 = 63 Volts

Now total potential at this point is given as

V = 45 - 60 + 63 = 48 Volts

3 0
3 years ago
A 78 kg skydiver can be modeled as a rectangular "box" with dimensions 24 cm × 35 cm × 170 cm . If he falls feet first, his drag
ANTONII [103]

Answer:

The terminal speed of his is 137.68 m/s.

Explanation:

Given that,

Mass of skydiver = 78 kg

Area of boxA =24\times35=840\ cm

Drag coefficient = 0.80

Density of air \rho= 1.2\times kg/m^3

We need to calculate the terminal velocity

Using formula of drag force

F_{d} = \dfrac{1}{2}\rho v^2Ac

Where,

\rho = density of air

A = area

C= coefficient of drag

Put the value into the formula

78\times9.8=\dfrac{1}{2}\times1.2\times v^2\times24\times10^{-2}\times35\times10^{-2}\times0.80

v^2=\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}

v=\sqrt{\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}}

v=137.68\ m/s

Hence, The terminal speed of his is 137.68 m/s.

3 0
3 years ago
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