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S_A_V [24]
3 years ago
6

A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a f

riction force of 15N. The block then slides up a frictionless ramp at 30 degrees reaching a 1.5m maximum height off the ground. What was the force constant of the spring
Physics
1 answer:
Setler [38]3 years ago
6 0

Answer:

7200 N/m

Explanation:

Metric unit conversion

100g = 0.1 kg

5 cm = 0.05 m

50 cm = 0.5 m

As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

W_f = F_fs = 15*0.5 = 7.5 J

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

E_p = mgh = 0.1*10*1.5 = 1.5 J

Therefore, as elastic energy is converted to potential energy and work of friction:

E_e = W_f + E_p

kx^2/2 = 7.5 + 1.5 = 9 J

k = 9*2/x^2 = 18/0.05^2 = 7200 N/m

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Ganezh [65]

Answer:

10.1 N

Explanation:

Your answer is 10.1 N, I don't actually know how to do it but I hope it helps.

8 0
3 years ago
While buying a hot plate you notice the resistance of the hot
marissa [1.9K]

Answer:

Current = 5.45amps

Power = 654 watts

Explanation:

E =120V

I =?

R = 22.02 Ohms

I= E/R

I= 120/22.02

I = 5.45AmPs

P = ?                    P= E x R

E = 120V              P= 120x5.45

I = 5.45 AmPs      P=  654W

7 0
2 years ago
( Can someone help? )
Murrr4er [49]

Answer:

Answer would be 0.33

Explanation:

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8 0
3 years ago
Read 2 more answers
which two gases in earths atmosphere are believed by scientists to be greenhouse gases that are major contributors to global war
Olenka [21]
Nitrogen and carbon dioxide??
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1 year 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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