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kiruha [24]
2 years ago
13

Relate 1 radian to 1 degree.how much is 90 degree in terms of radian?​

Physics
1 answer:
bogdanovich [222]2 years ago
3 0

The answer to the question is 1.5708~1.571.

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2. A person began running due east and covered 15 kilometers in 2.0 hours. What is the average velocity of the person?
Angelina_Jolie [31]

Answer:

7.5 km/h (2.1 m/s) due east

Explanation:

The average velocity of the person is given by:

v=\frac{d}{t}

where

d is the displacement

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In this problem,

d = 15 km is the displacement

t = 2.0 h is the time elapsed

so the average velocity is

v=\frac{15 km}{2.0 h}=7.5 km/h

and the direction is the same as the displacement (east).

We can also convert the velocity into SI units (m/s). We have:

d = 15 km = 15,000 m

t = 2.0 h * 3600 s/h = 7200 s

v=\frac{15,000 m}{7200 s}=2.1 m/s

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Answer:

Explanation:

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2 years ago
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tigry1 [53]
Spreads out in the medium, first choice
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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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