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mr Goodwill [35]
3 years ago
8

When cartographers represent the three-dimensional earth in two dimensions, what is likely to occur?

Physics
2 answers:
Harlamova29_29 [7]3 years ago
7 0
Distortion is likely to occur
Sladkaya [172]3 years ago
6 0

distortion is most likely to occur.

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Can you list the offensive position on a flag football team?
Alik [6]

Answer:

yes u can flag football has everything that pad football has so you can enlist on being offensive position but you have to play like you want that position

Explanation:

6 0
2 years ago
9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o
Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).

Now the force of friction is F_s=\mu*N, where N is the normal force and from the diagram it is F_y=mg*cos(\theta).

Thus F_s=\mu*N=\mu*mg*cos(\theta).

Therefore,

F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).

Substituting the value for F_H,m,\mu, and \:\theta we get:

F_{tot}= -38.63N.

Now acceleration is simply

a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.

The negative sign indicates that the acceleration is directed up the incline.

Part B:

d=\frac{1}{2} at^2

Which can be rearranged to solve for t:

t=\sqrt{\frac{2d}{a} }

Substitute the value of d=0.50m and a=7.7m/s and we get:

t=0.36s.

which is our answer.

Notice that in using the formula to calculate time we used the positive value of a, because for this formula absolute value is needed.

5 0
3 years ago
A 42.2 kg sled is pulled forward
zaharov [31]

The net force on the sledge  is 31.64N.

Frictional force = µkR

                         = 0.269 x 42.2 x 9.81 = 111.36

net force = 143N - 111.36N

               = 31.64N

refer  brainly.com/question/24557767

#SPJ2

     

7 0
2 years ago
A 35.4 kg girl is riding on the back of a 15.23 kg cart. the cart and the kid are both moving eastward at 4.25 m/s when she step
Grace [21]

Answer:

The final  velocity of the cart is  v_c = 7.02 \  m/s

Explanation:

From the question we are told that

    The mass of the girl is  m_g  = 35.4 \ kg

     The mass of the cart is  m_c  = 15.23 \ kg

      The speed of the cart and  kid(girl) is  v = 4.25 \ m/s

     The final velocity of  the girl is v_g  = 3.06 \  m/s

Let assume that velocity eastward is  positive and velocity westward is negative (Note that if we assume vise versa it wouldn't affect the answer )

   The total momentum of the system before she steps off the back of the cart

is mathematically evaluated as

        p__{T1}} = (m_g + m_c) * v

substituting values

        p__{T1}} = (35.4 + 15.23) * 4.25

        p__{T1}} =215.17 \  kg m /s

The total momentum after she steps off the back of the cart is mathematically evaluated as

        p__{T2}} = (m_g * v_g ) +(  m_c * v_c )

Where  v_c  is the final velocity of the cart

substituting values    

      p__{T2}} = (35.4 * 3.06 ) +(  15.23 * v_c )

       p__{T2}} = 108. 324 + 15.23  v_c

Now according to the law of conservation of momentum

       p__{T1}} =p__{T2}}

So  

       215.17 \  kg m /s =  108. 324 + 15.23  v_c

=>      v_c = 7.02 \  m/s

Since the value is positive it implies that the cart moved eastward

7 0
3 years ago
protons in an atomic nucleus are typically 10-15 m apart. what is the electric force of repulsion between nuclear protons?
Aleonysh [2.5K]

230 Newton

Electric charge consists of two types i.e. positively electric charge and negatively electric charge.There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :

F = electric force (N)

k = electric constant (N m² / C²)

q = electric charge (C)

r = distance between charges (m)

The value of k in a vacuum = 9 x 10⁹ (N m² / C²)

F = k(q1 q2)/ r^2

Distance between protons = d = 10⁻¹⁵ m

charge of proton = q = 1.6 × 10⁻¹⁹ C

Here q1=q2

electric force = F =230N

Coulomb's Law. Two protons in an atomic nucleus are typically separated by a distance of 2×10−15m. The electric repulsive force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart.

2 Nuclei and the Need for an Attractive Nuclear Force. The Coulomb force also acts within atomic nucleii, whose characteristic dimension is 10 m, which is called a fermi. There are two protons in a He nucleus, which repel each other because of the Coulomb force.

Find more about electric force of repulsion between nuclear protons

brainly.com/question/8404637

#SPJ4

5 0
1 year ago
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