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forsale [732]
2 years ago
10

What would you use to to measure the weight of a truck?? I’m needing a word that is 8 letters long

Physics
1 answer:
myrzilka [38]2 years ago
5 0

Answer:

Railroad Scale?

Explanation:

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State two applications of electrostatic
kaheart [24]
Atmospheric electricity and storms, electrostatic control filters, and industrial electrostatic seperation as well as spark discharge. these are just a few. hope it helps. 
6 0
3 years ago
Cual es el deporte que le da fortaleza y flexibilidad al cuerpo
AlexFokin [52]

Answer:

Aesthetic sports

Explanation:

Aesthetic sports are the one's that need well-developed physical qualities such as strength, agility, stamina, flexibility, and technical knowledge and artistry, in addition to technical ability and artistry. Elite athletes in these sports generally have a low abdominal fat , and the ranking is subjective.

In aesthetic sports like gymnastics, swimming, and figure skaters, dynamic and proactive flexibility is required.

4 0
2 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!
Aneli [31]

d^2=4^2+3^2\\\\d^2=16+9\\\\d^2=25\\\\d=5

3 0
3 years ago
A car starts from rest and accelerates to 14 m/s in 2 seconds. What was its acceleration
Romashka [77]

Answer:

7m/s^2

Explanation:

using v=u+at

since the car started from rest, u=0 , v=14m/s t=2s

a =acceleration.

14=0+a×2

14=0+2a

14=2a

a= 14/2 =7

a=7m/s^2

7 0
2 years ago
Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid
nataly862011 [7]

Answer:

A)

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B)

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load m_{LS} = 10000 kg

mass of flat bed m_{FB} = 20000 kg

initial speed of truck v_{0} = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs F_{N}     -------------let this be equation 1

where F_{N} = normal force = mg

so

Fs,max = μs mg

ma_{max} = μs mg

divide through by mass

a_{max} = μs g    ---------- let this be equation 2

in equation 2, we substitute in our values

a_{max} = 0.5 × 9.8 m/s²

a_{max} = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

v_{f}² = v_{0}² + 2aΔx

where v_{f} is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² /  9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

3 0
3 years ago
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