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irina [24]
3 years ago
12

Quanto vale a resultante de duas forças de mesmo módulo mesma direção e sentidos opostos atuando sobre um corpo de massa igual a

2 kg qual é o valor da aceleração desse corpo nessas condições
Physics
2 answers:
Crank3 years ago
6 0
A resultante das duas forças será zero, já q os sentidos são opostos e sãos iguais em módulo, elas se anulam. Logo, se a força resultante é zero, e F=ma, aceleração também será igual a zero.
Anton [14]3 years ago
6 0

Answer:

This question is in Portuguese, this means:

"How much is the result of two forces of the same module in the same direction and opposite ways acting on a body with a mass equal to 2 kg? what is the value of the acceleration of that body in these conditions?"

Now, i will answer in English.

When we are doing math with forces, we can use "Linear overlay", this means that we can add an subtract the forces between them.

Now, if you apply two forces of the same magnitude and same direction but in opposite ways (one is pointing in x positive and the other in x negative) we can directly add them:

F*(+x) + F(-x) = F(+x) - F(+x) = 0

Where (+x) and (-x) refer to the direction at which the forces are applied,

and it has sense, if you pull with the same force by opposite sides, the total force that the object suffers is zero.

And because the total force that the object suffers is zero, by the second Newton's equation f*m = a, the acceleration that the object suffers is also zero.

You might be interested in
A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block
Usimov [2.4K]

Answer:

a) v'=-0.227\ m.s^{-1}

b) v=1.36\ m.s^{-1}

Explanation:

Given:

mass of the lighter block, m'=0.02\kg

velocity of the lighter block, u'=0.68\ m.s^{-1}

mass of the heavier block, m=0.04\ kg

velocity of the heavier block, u=0\ m.s^{-1}

a)

Using conservation of linear momentum:

m'.u'+m.u=m'.v'+m.v

where:

v'= final velocity of the lighter block

v= final velocity of the heavier block

m'.u'=m'.v'+m.v

m'(u'-v')=m.v ........................(1)

Since kinetic energy is conserved in elastic collision:

\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2

m'(u'^2-v'^2)=m.v^2

m'(u'-v')(u'+v')= m.v^2

divide the above equation by eq. (1)

v=u'+v' .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

m'(u'-v')=m(u'+v')

\frac{m'+m}{m'-m} =\frac{u'}{v'}

\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}

v'=-0.227\ m.s^{-1} (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

m'(u'-v+u')=m.v

2m'.u'=(m-m')v

2\times 0.02\times 0.68=(0.04-0.02)\times v

v=1.36\ m.s^{-1}

6 0
3 years ago
A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

7 0
3 years ago
How high would you have to lift a 1000kg car to give it a potential energy of:
Elza [17]

Given parameters:

Mass of the car = 1000kg

Unknown:

Height  = ?

To find the heights for the different amount potential energy given, we need to understand what potential energy is.

Potential energy is the energy at rest due to the position of a body.

 It is mathematically expressed as:

          P.E  = mgh

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height of the car

Now the unknown is h, height and we make it the subject of the expression to make for easy calculation.

               h = \frac{P.E }{mg}

<u>For 2.0 x 10³ J;</u>

                  h  = \frac{2000}{1000 x 9.8}   = 0.204m

<u>For 2.0 x 10⁵ J;</u>

                  h  = \frac{200000}{9.8 x 1000}   = 20.4m

<u>For 1.0kJ  = 1 x 10³J; </u>

                  h  = \frac{1000}{9.8 x 1000}   = 0.102m

   

5 0
3 years ago
5. A wave with peaks separated by .34 m has a wavelength of ________________m.
Eduardwww [97]

Answer: 0.17 I think

Explanation:

I asked a doctor

8 0
3 years ago
While driving on the highway, you have a tire blowout. after you have slowed down and regained control, you should?
Svetllana [295]
When a person has experienced a tire blow out, there are steps to follow in order to prevent harm to others and self. After following the required and helpful steps after a tire blow out, slowed down and regain control, it is best that the person should head to a stop road or at the safe side of the lane where they won't be a bother to the road or to people driving in the high way. After pulling over to the side, it is advisable to turn on the emergency flashers of the car. This will set as a signal that you are in need of help or had gone through a problem.
4 0
3 years ago
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