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3 years ago

If the net force on an object is zero then the object has

Physics

1 answer:

Pavlova-9 [17]3 years ago

6 0

Answer:

If the net force is zero, then the body has an acceleration = 0 m/s²

Explanation:

Force: This is defined as the product of mass of a body and its acceleration. Force is a vector quantity because it can be represented both in magnitude and in direction. The S.I unit of force is Newton (N).

Net Force: This is the vector sum of all the the forces acting on a body.

Forces acting in the same direction are added while those acting in opposite directions are subtracted.

Example, if two forces F₁ and F₂ act on a body, in opposite direction, The net force F₃ is express below

F₃ = ma = F₁ - F₂ in the direction of F₁

From the example above, If the net force is zero, then the body has an acceleration = 0 m/s²

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Two sacks contain the same number of identical apples and are separated by a distance r. The two

andrezito [222]

Answer:

The appropriate response will be " $F_2=\frac{3}{4}F$ ". A further explanation is given below.

Explanation:

According to the question,

⇒ $F_1=\frac{G(m_1 m_2)}{r^2}$ ....(equation 1)

and,

⇒ $F_2=\frac{G(m_1 m_2)}{r^2}$ ....(equation 2)

Now,

On dividing both the equations, we get

⇒ $\frac{F_1}{F_2}=\frac{\frac{G(2)(2)}{(1)^2}}{\frac{G(1)(3)}{(1)^2}}$

⇒ $\frac{F_1}{F_2}= \frac{4}{3}$

⇒ $F_2=\frac{3}{4}F$

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3 years ago

A 0.119 kg ball is dropped from rest. if the magnitude of the ball's momentum is 0.817 kg·m/s just before it lands on the ground

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3 years ago

You are asked to design a horizontal curve with a 40-degree central angle (' = 40) for a two-lane road with 11-ft lanes. the des

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3 years ago

A 6.00 kg object is lifted vertically through a distance of 5.25 m by a light string under a tension of 80.0 N. Find: (2 marks)

jeka57 [31]

Explanation:

It is given that,

Mas of the object, m = 6 kg

It is lifted through a distance, h = 5.25 m

Tension in the string, T = 80 N

(a) By considering the free body diagram of the object, the forces can be equated as :

$T-mg=ma$

$a=\dfrac{T-mg}{m}$

$a=\dfrac{80-6\times 9.8}{6}$

$a=3.33\ m/s^2$

Work done by tension, $W_t=F\times h$

$W_t=80\times 5.25$

$W_t=420\ J$

(b) Work done by gravity, $W_g=mgh$

$W_g=6\times 9.8\times 5.25$

$W_g=308.7\ J$

(c) Let v is the final speed of the object and u = 0

$v=\sqrt{2ah}$

$v=\sqrt{2\times 3.33\times 5.25}$

v = 5.91 m/s

Hence, this is the required solution.

7 0

4 years ago

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)