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never [62]
4 years ago
12

The Force of friction between an object and the surface upon which it is sliding is 12N. The weight of the object is 20N. What i

s the coefficient of kinetic friction?
Use equation Ffr=Uk(mg)
Physics
1 answer:
VashaNatasha [74]4 years ago
6 0

Answer:

Explanation:

Reducing Sliding Friction. You can reduce the resistive force of sliding friction by applying lubrication between the two surfaces in contact, by using rollers, or by decreasing the normal force

You might be interested in
It took 3.5 hours for a train to travel the distance between two cities at a velocity of 120mi/h. How many miles lie between the
erma4kov [3.2K]

Answer:

420 miles

Explanation:

3.5 * 120 mi/h = 420 miles

3 0
3 years ago
What times what equals 400
Radda [10]

If you're willing to consider fractions or decimals,
then there are an infinite number of answers. 
Like (2.5 x 160), and (15 x 26-2/3).

If you want to stick to only whole numbers,
then these 8 combinations do:

1, 400
2, 200
4, 100
5, 80
8, 50
10, 40
16, 25
20, 20

7 0
4 years ago
A particle initially at rest moves along in a line so that its acceleration is a(t) = 10/(t+1) for t>=0.
ddd [48]

Answer: 16.09m/s

Explanation:

Given the acceleration a(t) = 10/(t+1), to derive the velocity function, we need integrate the acceleration function.

v(t) = integral{10/t+1}dt

Since all constants always come out of the integral, the equation becomes;

v(t) = 10integral{1/t+1}dt

One of the integral law is that if the numerator of the function to be integrated is the differential of the denominator, the resulting answer will be natural logarithm of the denominator i.e ln(t+1) since the denominator is ln(t+1) and if differentiated will give us 1 which is the numerator hence, the reason for the answer ln(t+1)

v(t) = 10ln(t+1)

@ t= 4, the velocity of the particle

v(4) = 10ln(4+1)

v(4) = 10ln5

v(4) = 16.09m/s²

Therefore, the velocity of the particle at time t=4 is 16.09m/s

4 0
3 years ago
6th grade science please help​
katovenus [111]
The answer is convention the last one
6 0
3 years ago
Read 2 more answers
In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance
maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = X_{L} = j\omega L = 2\pi fL

where

R = resistance

X_{L} = Inductive Reactance

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

C = \frac{\epsilon_{o}A}{x}

where

x = separation between the parallel plates

Thus

C ∝ \frac{1}{x}

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

Also,

Z ∝ I

Therefore,

\frac{Z}{I} = \frac{Z'}{I'}

\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}

{R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})

{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})

Solving the above eqn:

X_{C} = 4R

6 0
3 years ago
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