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mariarad [96]
4 years ago
5

What's the formula for acceleration when force is included.

Physics
2 answers:
Bezzdna [24]4 years ago
4 0
Resultant force (newton, N) = mass (kg) × acceleration (m/s2<span>)
Hope this helped :)</span>
Dima020 [189]4 years ago
3 0

Acceleration of an object  = 

         (the net force acting on it)
divided by
         (its mass)    .
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Crest is the answer
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3 years ago
What is the energy of a photon with a frequency of 1.7 × 1017 Hz? Planck’s constant is 6.63 × 10–34 J•s.
Karo-lina-s [1.5K]
The energy carried by a single photon is given by
E=hf
where h is the Planck's constant and f is the frequency of the photon.

The photon of our exercise has a frequency of f=1.7 \cdot 10^{17} Hz, therefore its energy is
E=hf=(6.63 \cdot 10^{-34}Js)(1.7 \cdot 10^{17} Hz)=1.1 \cdot 10^{-16} J
3 0
4 years ago
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Graph the following data on the graph, then use the graph to determine the half-life of this isotope.
Harlamova29_29 [7]

Answer:

4, 56

Explanation:

Hope this helped!

7 0
3 years ago
Water flows along a streamline down a river of constant width. Over a short distance, the water slows from speed v to v/3. Which
kvasek [131]

Answer:

a. It became deeper by a factor of 3.

Explanation:

What we have is water flowing down a river with constant width. The water slows from speed v to v3 over a shirt distance

Using the equation of continuity

A1V1 = A2V2 ----1

A1 is the area of rectangle

V1 is the velocity of water

Area of rectangle = length x width

We rewrite equation 1 as

λ1w1v1 = λ2w2v2

We have w1 = w2

λ1v1 = λ2v2

λ1*v1 = λ2*v/3

λ1 = λ2/3

So it becomes deeper by a factor of 3

8 0
3 years ago
A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the
rjkz [21]

Answer:

a) F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}

b) \mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

\Sigma F_{y}=0

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

-F_{y}-W+N=0

When solving for the normal force we get:

N=F_{y}+W

we know that

W=mg

and

F_{y}=Fsin \theta

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

f_{k}=\mu_{k}N

so we can find the kinetic friction by substituting for N, so we get:

f_{k}=\mu_{k}(F sin \theta +mg)

Now we can find the sum of forces in x:

\Sigma F_{x}=0

so after analyzing the diagram we can build our sum of forces to be:

-f+F_{x}=0

we know that:

F_{x}=Fcos \theta

so we can substitute the equations we already have in the sum of forces on x so we get:

-\mu_{k}(F sin \theta +mg)+Fcos \theta=0

so now we can solve for the force, we start by distributing \mu_{k} so we get:

-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0

we add \mu_{k}mg to both sides so we get:

-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg

Nos we factor F so we get:

F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg

and now we divide both sides of the equation into (cos \theta-\mu_{k} sin \theta) so we get:

F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

f_{s}=\mu_{s}(F sin \theta +mg)

and

F cos θ = f

when substituting one into the other we get:

F cos \theta=\mu_{s}(F sin \theta +mg)

which can be solved for the static friction coefficient so we get:

\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}

which is the answer to part b.

3 0
4 years ago
Read 2 more answers
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