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8090 [49]
3 years ago
13

How many pounds of force must a guardrail’s top rail be able to withstand? A. 150 lbs. B. 200 lbs. C. 400 lbs. D. 5,000 lbs.

Physics
2 answers:
oksian1 [2.3K]3 years ago
7 0
It has to be atleast 300 lbs which is b
Greeley [361]3 years ago
3 0
At least 200 pounds.
hope this one helps
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The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If
mart [117]

Answer:  29.50 m

Explanation: In order to calculate the higher accelation to stop a train  without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction  multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m

5 0
2 years ago
1.A Radio station broadcasts modern song on medium wave 350 Hz every day at ten o’clock in the morning. The velocity of radio wa
love history [14]

Answer:

ans \:  = \boxed{{4.8 \times 10}^{ - 4}  Hz}

Explanation:

given \to \\  f_{r} = 350 \:  \\ v_{r} =  {3 \times 10}^{8}  \\ but \to \\ v = f \gamma   \to \:  \gamma  =  \frac{v}{f}  : hence \to \\  \gamma _{r} =  \frac{v_{r}}{f_{r}}   =  \frac{3 \times 10^{8} }{350}   =  \boxed{857,142.85714 \: m}\\ therefore \to \\ given \to \\  f_{w} = water \: frequency = \:  \boxed{  ?}\:  \\ v_{w} =  14 50 \\ but \to \\ v = f \gamma   \to \:  \gamma  =  \frac{v}{f}  : hence \to \\  \gamma _{w} =  \frac{v_{w}}{f_{w}}   =  \frac{1}{100}  \times \gamma _{r}  =  \frac{1}{100}  \times 857,142.85714  \\\gamma _{w}  =  \boxed{8,571.4285714 \: m} : hence \to \:  \\ f_{w} =  \frac{v_{w}}{ \gamma _{w}}  =  \frac{1450}{8,571.4285714}  =  \boxed{0.1691666667} \\ if \: the \: number \: of \: times = \boxed{ x} \\ f_{r} (x)=f_{w} \\ (x) =  \frac{f_{w}}{f_{r}}  =  \frac{0.1691666667}{350}  = 0.0004833333 \\ hence \to \\ the  \: frequency  \: of \:  the \:  radio  \: wave  \: is \to \:   \boxed{{4.8 \times 10}^{ - 4}  }\:  \\ that  \: of  \: the \:  wave  \: created  \: in  \: the  \: water.

♨Rage♨

8 0
3 years ago
A sound wave moving with a speed of 1500 m/s is sent from a submarine to the ocean floor. It reflects off the
Alekssandra [29.7K]

Answer:

the distance between the submarine and the ocean floor is 11,250 m

Explanation:

Given;

speed of the wave, v = 1500 m/s

time of motion of the wave, t = 15 s

The time taken to receive the echo is calculated as;

time \ of \ motion \ (t) = \frac{total \ distance }{speed \ of \ wave} = \frac{2d}{v}  \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{1500 \times 15}{2} \\\\d = 11,250 \ m

Therefore, the distance between the submarine and the ocean floor is 11,250 m

3 0
2 years ago
Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.
Katen [24]
You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>
5 0
3 years ago
According to newton's first law, what is required to make an object slow down?
Luden [163]
I believe it is friction

3 0
2 years ago
Read 2 more answers
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