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emmainna [20.7K]
1 year ago
14

What are close-toed shoes least likely to provide protection against?

Physics
2 answers:
Sphinxa [80]1 year ago
7 0

Hmm.. Well i would say A or C because water and acid would just soak threw it

In-s [12.5K]1 year ago
7 0

hydrogen gas produced during a reaction

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Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n
Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

<u>FOR FRIEND ON MOON</u>:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

<u>FOR ME ON EARTH</u>:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

m(earth) = 0.1 kg

Since, the mass of gold on moon is greater than the mass of moon on earth.

<u>Therefore, the friend on moon will be richer.</u>

7 0
2 years ago
The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2
masha68 [24]

Answer:

the normal stress induced by  the concrete post \sigma_c = 67.26 psi

the normal stress induced by the steel \sigma_s = - 1795.84 psi

Explanation:

Given that:

Modulus for elasticity for concrete post E_c = 3.6 *10^6 psi

Thermal coefficient for concrete post  \alpha _c = 5.5 *10^{-6}/^0F

Modulus for elasticity of steel bar E_s = 29*10^6psi

Thermal coefficient of steel bar \alpha _2 = 6.5*10^{-6}/^0F

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod A_s = 6(\frac{ \pi}{4} )(d_s)^2

= 6(\frac{ \pi}{4} )(\frac{7}{8} )^2

= 3.61 in²

Area of concrete parts A_c = (10)(10) - A_s

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= [\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)

= [\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)

= [(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}

= 1.234*10^{-8}P = 8.0*10^{-5}

P = \frac {8.0*10^{-5}}{1.234*10^{-8}}

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

\sigma_c =\frac{P}{A_c}

\sigma_c = \frac{6482.98}{96.39}

\sigma_c = 67.26 psi

Thus, the normal stress induced by  the concrete post \sigma_c = 67.26 psi

Also; the normal stress in the steel bars  induced as a result of temperature rise is as follows:

\sigma_s = \frac{-P}{A_s}

\sigma_s =\frac{-6482.98}{3.61}

\sigma_s = - 1795.84 psi

Thus, the normal stress induced by the steel \sigma_s = - 1795.84 psi

6 0
2 years ago
An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o
8_murik_8 [283]

Answer:

The gplanet is 0.193 m/s^2

Explanation:

The speed of the pulse is:

v=\frac{lengthofthewipe}{traveltime} =\frac{1.6}{0.0656} =15.24m/s

v=\sqrt{\frac{MgL}{m} } \\v^{2} =\frac{MgL}{m} \\g=\frac{mv^{2} }{ML}

where

m=mass of the wire=4 g= 4x10^-3 kg

M=mass of the object= 3 kg

Replacing values:

g=\frac{4x10^{-3}*15.24^{2}  }{3*1.6} =0.193 m/s^{2}

7 0
2 years ago
A radar used to detect the presence of aircraft receives a pulse that has reflected off an object 5 ✕ 10−5 s after it was transm
Sunny_sXe [5.5K]

Answer:

7500 m

Explanation:

The radar emits an electromagnetic wave that travels towards the object and then it is reflected back to the radar.

We can call L the distance between the radar and the object; this means that the electromagnetic wave travels twice this distance, so

d = 2L

In a time of

t=5\cdot 10^{-5}s

Electromagnetic waves travel in a vacuum at the speed of light, which is equal to

c=3.0\cdot 10^8 m/s

Since the electromagnetic wave travels with constant speed, we can use the equation for uniform motion ,so:

d=vt (1)

where

v=c=3.0\cdot 10^8 m/s

t=5\cdot 10^{-5}s

d=2L, where L is the distance between the radar and the object

Re-arranging eq(1) and substituting, we find L:

L=\frac{vt}{2}=\frac{(3.0\cdot 10^8)(5\cdot 10^{-5})}{2}=7500 m

7 0
2 years ago
A runner runs 4875 ft in 6.85 minutes. what is the runners average speed in miles per hour?
yanalaym [24]

The average speed can be easily calculated by taking the ratio of distance and time. That is:

average speed = distance / time

 

so calculating:

average speed = 4875 ft / 6.85 minutes

<span>average speed = 711.68 ft / min</span>

8 0
2 years ago
Read 2 more answers
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