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drek231 [11]
3 years ago
9

Once you start pulling your object with less force than friction, what should you expect your object to do? What about when your

object is pulled with more force than friction?
NO LINK S​
Physics
2 answers:
forsale [732]3 years ago
8 0

#Case -1

If Pulling force is less than frictional force the object won't move .

#Case-2

If Pulling force is greater than frictional force then object will be .

In order to calculate friction force you need Limiting friction first .

\\ \sf\longmapsto F_L=\mu sN

u s is coefficient of static friction and N is normal reaction

Or

\\ \sf\longmapsto F_L=\mu smg

  • As N=mg
GenaCL600 [577]3 years ago
5 0

Answer:

#Case -1

If Pulling force is less than frictional force the object won't move .

#Case-2

If Pulling force is greater than frictional force then object will be .

In order to calculate friction force you need Limiting friction first .

\begin{gathered}\\ \sf\longmapsto F_L=\mu sN\end{gathered}

⟼F

L

=μsN

u s is coefficient of static friction and N is normal reaction

Or

\begin{gathered}\\ \sf\longmapsto F_L=\mu smg\end{gathered}

⟼F

L

=μsmg

As N=mg

Explanation:

Sry......

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Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and
djverab [1.8K]

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Here we are given,

k = The spring has a spring constant.

= 25.5 N/m.

x_f = Final length of the spring .

= \sqrt{1.25^2+1.8^2}  -0.60

= 1.591 m

x_i= The initial length of the spring.

= 1.25−0.60

=0.65 m

y=The collar then travels downward a distance.

=  1.80 m.

m= The mass of the collar.

=3.55 kg

v_c = the velocity of the collar.

= 3.39 m/s.

g = The acceleration due to gravity.

= 9.81 m/s²

We have to calculate the magnitude of the average friction force exerted on the collar = F

Now we put the known values in the above equation, we get;

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Or, \frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2}  = 3.55\times 9.81\times 1.80

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

Learn more about friction:

brainly.com/question/24338873

#SPJ4

Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .

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2 years ago
A small rectangular block of metal must be a magnet is it
Arlecino [84]
D. Is repelled by a magnet
3 0
4 years ago
What statements correctly describe theories
elixir [45]

Theories result from several repeated experiments.

Theories explain observations and hypotheses.

Theories may be revised over time.

Explanation:

Scientific theories are purely explanations into an observation and hypothesis. The are general binding explanations that have been developed from several tests.

  • Theories are products of different stages of experiments in their own regard.
  • For a theory to be accepted by the scientific community, its hypothesis statement must be:
  1. Testable
  2. Repeated
  3. Falsifiable
  • Based on new evidence, a theory may be revised with time. One of such is the Dalton's atomic theory with a modern atomic theory version now.

Learn more:

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8 0
3 years ago
M=3000km v=25m/s what’s the momentum
valina [46]
<h3>Answer:</h3>

Momentum of the given body will be : 75000 Kg m/s

<h3>Explanation:</h3>

According to Newton's first law of motion, all bodies continue to be in the state of rest or motion unless an external force is applied on the body. We can use this in the case of momentum also

The formula of momentum is given by :

:\implies \sf\quad \sf \:  P = mv

Here, we are given the mass of the body ( m ) as 3000kg and the velocity of the body ( v ) as 25 m/s. On putting the values in the formula:

\begin{aligned}&:\implies \sf\quad \sf \:  P = mv \\& :\implies \sf\quad \sf \:  P = 3000 \times 25 \\ & :\implies \sf\quad \sf \:   \boxed{ \sf \: P = 75000kgm {s}^{ - 1} } \end{aligned}

Momentum is associated with the mass of the moving body and can be defined as the quantity of motion measured as a product of mass and velocity.

8 0
2 years ago
To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv
Lesechka [4]

Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

        E = k λ (-1/x){(-1/x)}^{d+L} _{d}

   

Evaluating

        E = k λ [ 1/d  - 1/ (d+L)]

Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

3 0
3 years ago
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