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ZanzabumX [31]
2 years ago
10

Which factors are always part of linear motion?

Physics
2 answers:
Daniel [21]2 years ago
4 0

Answer:

Explanation:

In linear motion, the directions of all the vectors describing the system are equal and constant which means the objects move along the same axis and do not change direction.

Paha777 [63]2 years ago
3 0
I’m linear motion, the directions of all the vectors describing the system are equal and constant which means the object move along the same axis and do not change directions

Hope this helped
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A fish swimming in a horizontal plane has velocity i = (4.00 + 1.00 ) m/s at a point in the ocean where the position relative to
ryzh [129]

Answer:

a. The horizontal component of acceleration a₁ = 0.68 m/s²

The vertical component of acceleration a₂ = -0.11 m/s²

b. -9.19° = 350.81° from the the positive x-axis

Explanation:

The initial velocity v₁ of the fish is v₁ = 4.00i + 1.00j m/s. Its final velocity after accelerating for t = 19.0 s is v₂ =  17.0i - 1.00j m/s

a. The acceleration a = (v₂ - v₁)/t = [17.0i - 1.00j - (4.00i + 1.00j)]/19 = [(17.0 -4.0)i - (-1.0 -1.0)j]/19 = (13.0i - 2.0j)/19 = 0.68i - 0.11j m/s²

The horizontal component of acceleration a₁ = 0.68 m/s²

The vertical component of acceleration a₂ = -0.11 m/s²

b. The direction of the acceleration relative to the unit vector i,

tanθ = a₂/a₁ = -0.11/0.68 = -0.1618

θ = tan⁻¹(-0.1618) = -9.19° ⇒ 360 + (-9.19) = 350.81° from the the positive x-axis

6 0
3 years ago
If 1 foot is 30.28 centimeters.how many cm is 130 feet
uranmaximum [27]

Answer:3,936.4

Explanation:

If 1 foot is 30.28 centimeters multiply 30.28 by 130

6 0
3 years ago
A pendulum is set in motion and time. The time measured for 20 complete swings is 308.
laiz [17]

Answer:

<em><u>solution</u></em>

<em>3</em><em>0</em><em>8</em><em>=</em><em>2</em><em>0</em><em> </em><em>swings</em><em> </em>

<em> </em><em>?</em><em>:</em><em>:</em><em>:</em><em> </em><em>=</em><em>1</em>

<em>(</em><em> </em><em>3</em><em>0</em><em>8</em><em>×</em><em>1</em><em>)</em><em>÷</em><em>2</em><em>0</em>

<em>3</em><em>0</em><em>8</em><em>÷</em><em>2</em><em>0</em>

<em>1</em><em>5</em><em>4</em><em>÷</em><em>1</em><em>0</em>

<em>=</em><em>1</em><em>5</em><em>.</em><em>4</em>

<em>=</em>15.4

6 0
3 years ago
Suppose the electrons and protons in 1g of hydrogen could be separated and placed on the earth and the moon, respectively. Compa
MAXImum [283]

Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}

Where:

G= 6.674×10^−11 N · (m/kg)2

k =  8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}   --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is  5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}

Therefore

\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field

7 0
3 years ago
A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.
pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t  = 6.55 \ s

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

8 0
3 years ago
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