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3 years ago

How does the direction of friction relate to the direction of motion

1 answer:

5 0

It reacts to the motion. If the mass hanging from the pulley was overwhelmingly heavier than the mass on the ramp, it'll obviously pull the ramp mass up and thus friction would be trying to oppose this and vice versa.

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HELPPP why is the united states considered energy rich, while japan is considered energy poor

Temka [501]

Answer:

Energy resources can be measured. They will include the fossil fuels, geothermal and hydroelectric potential, and increasingly the renewable resources. When the US list is compared to the World it is considered energy Rich. When Japan's list is compared to the world standard it considered energy poor.

A changing technology like nuclear fusion could substantially change the assessment.

Japan does not have any substantial, oil, coal, gas, deposits, while the US does.

Explanation:

6 0

3 years ago

Hi can someon help me how to answer this? Btw I'm from Philippines

arlik [135]

Answer:

Test 1

1.True

2.True

3.True

4.False

5.True

6.True

7.False

8.True

9.True

10.True

yung iba nasa pic

4 0

2 years ago

In outer space a rock of mass 12 kg is acted on by a constant net force 21, −11, 44 N during a 4 s time interval. At the end of

Luda [366]

Answer:

(100,84.67,89.33) m/s

Explanation:

From Newton's second law,

Impulse = Final momentum- initial momentum.

I = Mf-Mi.............. Equation 1

make Mi the subject of the equation

Mi = Mf-I............ Equation 2

Where I = Impulse. Mf = final momentum, Mi = initial momentum.

But

I = Ft.................... Equation 3

Where F = Force, T = time

Given: F = (21,-11,44) N, t = 4 s.

I = 4(21,-11,44)

I = (84,-44,176) N.s

Also,

Mf = mVf................ Equation 4

Where m = mass, Vf = Final velocity

Given: m = 12 g, Vf = (107,81,104) m/s.

Substitute into equation 4

Mf = 12(107,81,104)

Mf = (1284,972,1248) kgm/s.

Substitute the value of I and Mf into equation 2

Mi = (1284,972,1248)-(84,-44,176)

Mi = (1200,1016,1072) kgm/s

Also,

Mi = mVi

Where Vi = Initial velocity.

make Vi The subject of the equation

Vi = Mi/m.......................... Equation 5

Given: m = 12 kg, Mi = (1200,1016,1072) kgm/s

Substitute into equation 5

Vi = (1200,1016,1072)/12

Vi = (100,84.67,89.33) m/s

6 0

3 years ago

an iron rod of length 100m at 10 degree Celsius is used to measure a distance of 2km on a day when the temperature is 40 degree

german

Answer:

0.68 m

Explanation:

α = dL / L1*(dT)

dL = L1(dT) * α

Initial length, L1 = 100

Chang in Length =dL

α linear expansivity ; dL = change in length ; dT = change in temperature ; L1 = initial length

α of iron rod = 1.13 * 10^-5 k

dL = 100(40 - 10) * 1.13 * 10^-5

dL = 100(30) * 1.13 * 10^-5

dL = 3000 * 1.13 * 10^-5

dL = 3390 * 10^-5

dL = 0.0339 m

Error :

Distance measured = 2km = (2 * 1000) = 2000m

[Distance measured / (initial length + change in length)] × change in length

Error = (2000 / (100 + 0.0339)) * 0.0339

Error = (2000 / 100.0339) * 0.0339

Error = 19.993222 * 0.0339

Error = 0.6777702

Error = 0.68 m

4 0

3 years ago

A child of mass M is swinging on a swing set. The ropes attaching the swing to the top bar have length L. Find the gravitational

myrzilka [38]

Answer:

(a) 0

(b) 10ML

(d) $10ML(1 + sin(\phi))$

Explanation:

(a) When hanging straight down. The child is at the lowest position. His potential energy with respect to this point would also be 0.

(b) Since the rope has length L m. When the rope is horizontal, he is at L (m) high with respect to the lowest swinging position. His potential energy with respect to this point should be

$E_h = mgh = 10ML$

where g = 10m/s2 is the gravitational acceleration.

(c) At angle $\theta$ from the vertical. Vertically speaking, the child should be at a distance of $Lcos(\theta)$ to the swinging point, and a vertical distance of $L - Lcos(\theta)$ to the lowest position. His potential energy to this point would be:

$E_{\theta} = mgh = 10M(L - Lcos(\theta)) = 10ML(1 - cos(\theta))$

(d) at angle $\phi$ from the horizontal. Suppose he is higher than the horizontal line. This would mean he's at a vertical distance of $Lsin(\phi)$ from the swinging point and higher than it. Therefore his vertical distance to the lowest point is $L + Lsin(\phi) = L(1 + sin(\phi))$

His potential energy to his point would be:

$E_{\phi} = mgh = 10ML(1 + sin(\phi))$

5 0

3 years ago