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marysya [2.9K]
3 years ago
15

Need Help on this image is below thanks.....

Physics
2 answers:
Alik [6]3 years ago
8 0
The answer is b.mass
umka2103 [35]3 years ago
7 0

Answer:

Mass , this is the answer because mass allows objects of any size to be moved depending on the force of the impact , push or turn

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There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the cor
VMariaS [17]

Answer:

 F = 7.68 10¹¹ N,  θ = 45º

Explanation:

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

          F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

          cos 45 = F₂₄ₓ / F₂₄

          sin 45 = F_{24y) / F₂₄

          F₂₄ₓ = F₂₄ cos 45

          F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

          Fₓ = -F₂₁ + F₂₄ₓ

          Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis  

         F_y = - F₂₃ + F_{24y}

         F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

         F₂₁ = F₂₃

Let's use Coulomb's law

         F₂₁ = k q₁ q₂ / r₁₂²

       

the distance between the two charges is

         r = a

         F₂₁ = k q² / a²

we calculate F₂₄

           F₂₄ = k q₂ q₄ / r₂₄²

the distance is

           r² = a² + a²

           r² = 2 a²

         

we substitute

           F₂₄ = k  q² / 2 a²

we substitute in the components of the forces

          Fx = - k \frac{q^2}{a^2} +  k \frac{q^2}{2 a^2}  \ cos 45

          Fx = k \frac{q^2}{a^2}  ( -1 + ½ cos 45)

          F_y = k \frac{q^2}{a^2} ( -1 +  ½ sin 45)    

         

We calculate

            F₀ = 9 10⁹ 4.25² / 0.440²

            F₀ = 8.40 10¹¹ N

       

            Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

            Fₓ = -5.43 10¹¹ N

         

remember cos 45 = sin 45

             F_y = - 5.43 10¹¹  N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

          F = -5.43 10¹¹ (i + j)   N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

          F = \sqrt{F_x^2 + F_y^2}

          F = 5.43 10¹¹  √2

          F = 7.68 10¹¹ N

in angle is

           θ = 45º

7 0
3 years ago
On the asteroid Cere, the acceleration due to gravity is about 0.27 m/s^2. How long would it take a ball to fall to the ground i
KiRa [710]

Answer:

F. 25.82 s

Explanation:

Given:

Δy = 90 m

v₀ = 0 m/s

a = 0.27 m/s²

Find: t

Δy = v₀ t + ½ at²

90 m = (0 m/s) t + ½ (0.27 m/s²) t²

t = 25.82 s

7 0
4 years ago
A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car?
aksik [14]

ANSWER


C. F=4.5 \times10^3. newtons


EXPLANATION


According to Newton's second law,



F_{net}=ma, where



m=1.00\times 10^3kg is the mass measured in kilograms.


and


a=4.5ms^{2} is the acceleration in metres per second square.



We substitute these values to obtain,


F=1.00\times10^3 \times 4.5.



We rearrange to get,


F=1.00\times4.5 \times10^3.


We multiply out the first two numbers and leave our answer in standard form to get,



F=4.5 \times10^3 N.



The correct answer is C


3 0
4 years ago
X-rays with frequency 3 ⨉ 10 18 Hz shine on a crystal, producing an interference pattern. If the first bright spot is observed a
FrozenT [24]

Answer:

5.1645\times 10^{-11}\ m

Explanation:

n = Order = 1

c = Speed of light = 3\times 10^8\ m/s

f = Frequency = 3\times 10^{18}\ Hz

\theta = Angle = 75.5^{\circ}

Lattice spacing is given by

d=\dfrac{n\lambda}{2\sin\theta}\\\Rightarrow d=\dfrac{n\times c}{f2\sin\theta}\\\Rightarrow d=\dfrac{1\times 3\times 10^{8}}{3\times 10^{18}\times 2\times \sin75.5^{\circ}}\\\Rightarrow d=5.1645\times 10^{-11}\ m

The lattice spacing of the crystal is 5.1645\times 10^{-11}\ m

6 0
3 years ago
Si usted trata de contrabandear lingotes de oro llenando su mochila, cuyas dimensiones son de 56 cm * 28 cm * 22 cm, ¿cuál sería
kati45 [8]

Answer:

Mass, m = 665.77 kg

Explanation:

The question says that "If you try to smuggle gold bars by filling your backpack, whose dimensions are 56 cm * 28 cm * 22 cm, what would be its mass? "

The dimension of gold bar is 56 cm×28 cm×22 cm

We know that the density of gold is 19.3 g/cm³

The mas per unit volume of a material is called its density. It can be given by :

d=\dfrac{m}{V}\\\\m=d\times V\\\\m=(19.3\ g/cm^3)\times (56\times 28\times 22\ cm^3)\\\\m=665772.8\ g\\\\m=665.77\ kg

So, the mass of the gold bar is 665.77 kg.

5 0
3 years ago
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