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Romashka-Z-Leto [24]
3 years ago
13

12)

Physics
1 answer:
Nataliya [291]3 years ago
5 0

Answer: b

Explanation: the two pieces will repel as both have obtained a static charge.

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the layers of atmosphere are heated through radiation and convection.

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2 years ago
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Is the frictional force the same as the applied force when the net force equals zero?
DerKrebs [107]

Answer:

Since the net force is to the right (in the direction of the applied force), then the applied force must be greater than the friction force. The friction force can be determined using an understanding of net force as the vector sum of all the forces.

Explanation:

6 0
2 years ago
A 12.65 g sample of a radioactive substance is allowed to decay for 17.22 min. At that time, the sample weighed 3.115 g. What is
Mademuasel [1]

Answer:

8.61 min

Explanation:

original mass= 12.65

first half life = 12.65/2 = 6.325

second half life = 6.325/2 = 3.1625

Note : 3.1625 is the closest to the value (3.115) given so we work with it

total time for decay =17.22

therefore two decays = 17.22/2= 8.61

8 0
3 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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