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andrew-mc [135]
3 years ago
6

Newton's Three laws talk about all of the following:

Physics
1 answer:
Mrrafil [7]3 years ago
6 0
Force, because all three laws talk about how an object needs to be acted upon to move (an object in motion will stay in motion, an object at rest will stay at rest, every action has an equal opposite reaction)
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In the case of a swinging pendulum, Potential Energy is greatest when ____.
GREYUIT [131]
I am almost 100% positive the answer is A.
4 0
3 years ago
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Two arrows are fired horizontally with the same speed of
Fed [463]

Answer:

Explanation:

Given

mass of each arrow=0.1 kg

velocity of arrow=30 m/s

One arrow is fired u=due to east and another towards south

Momentum of first arrow

P_1=0.1\left ( 30\hat{i}\right )=3\hat{i}

P_2=0.1\left ( -30\hat{j}\right )=-3\hat{j}

Total momentum P

P=P_1+P_2

P=3\hat{i}-3\hat{j}

magnitude |P|=\sqrt{3^2+3^2}=3\sqrt{2}

direction

tan\theta =\frac{-3}{3}

\theta =45^{\circ} clockwise w.r.t to east

6 0
3 years ago
Determine the magnitude of the resultant force and its direction using both the parallelogram and Cartesian vector notation meth
Alika [10]

Answer:

   F = 1494.52 N,   θ = 44º

Explanation:

For the sum of vectors by the parallelogram method, see attached, the vectors are drawn, the parallelogram is completed and a vector is drawn from the origin of the two vectors to the end point of the rectangle, this is the resulting vector.

The attachment shows this roughly.

For the Cartesian coordinate method, each vector is decomposed into its components, they are added algebraically and then the resulting vector is composed in the form of a module and angles

we use trigonometry to decompose the vectors.

The coordinate system can be seen in the attachment

           sin θ = y / R

           cos θ = x / R

            y = R sin θ

            x = R cos θ

Vector 1

module F₁ and angle β₁ = 50

            sin 50 = \frac{F_{1y} }{F_1}

            cos 50 = \frac{F_{1x} }{F_1}

            F_{1y} = F₁ sin 50

            F₁ₓ = F₁ cos 50

            F_{1y} = 600 sin 50 = 459.63 N

            F₁ₓ = 600 cos 50 = 385.67 N

Vector 2

modulus F₂ = 900N, angle β₂ = 40

            F_{2y} = 900 sin 40 = 578.51 N

            F₂ₓ = 900 cos 40 = 689.44 N

we find the resultant of each component

           F_{y} =F_{1y} + F_{2y}

           F_{y}  = 459.63 + 578.51

           F_{y}  = 1038.14 N

 

            Fₓ = F₁ₓ + F₂ₓ

            Fₓ = 385.67 + 689.44

             Fₓ = 1075.11 N

We use the Pythagorean theorem to find the modulus of the resultant

            F = Fₓ² + F_{y}^2

            F = √(1075.11² + 1038.14²)

            F = 1494.52 N

we use trigonometry for the angle

            tan θ = F_y / Fₓ

            θ = tan⁻¹ (F_y / Fₓ)

            θ = tan⁻¹ (1038.14 / 1075.11)

            θ = 44º

8 0
3 years ago
Three tiny charged metal balls are arranged on a straight line. The middle ball is positively charged and the two outside balls
Dmitrij [34]

Answer:

(a) 189.23 N, (b) 47.31 N and (c) 141.92 N.

Explanation:

Three balls are shown in figure having charge q=1.45 \mu C. The middle ball, B, is positively charged having charge +q, and the remaining two outside balls, A and C, are negatively charged having charged -q as shown.

AC=20 cm and AB=BC=10 cm as B is the mid-point of AC.

Let d_1=AC=20\times 10^{-3}m and d_2=AB=BC=10\times 10^{-3}m

From Coulomb's law, the magnitude of the force, F, between two point charges having magnitudes q_1 \& q_2, separated by distance, d, is

F=\frac {1}{4\pi\epsilon_0}\frac {q_1q_2}{d^2}\;\cdots (i)

where, \epsilon_0 is the permittivity of free space and

\frac {1}{4\pi\epsilon_0}=9\times 10^9 in SI units.

This force is repulsive for the same nature of charges and attractive for the different nature of charges.

Now, Using equations(i),

(a) The magnitude of attraction force between balls A and B is

F_{AB}=F_{BC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_2)^2}

\Rightarrow F_{AB}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(10\times 10^{-3}\right)^2}

\Rightarrow F_{AB}=189.23 N

(a) The magnitude of the repulsive force between balls A and C is

F_{AC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_1)^2}

\Rightarrow F_{AC}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(20\times 10^{-3}\right)^2}

\Rightarrow F_{AC}=47.31 N

(c) The magnitude of the net force, F_{net}, on the outside of the ball is,

F_{net}=189.23-47.31 N

\Rightarrow F_{net}=141.92 N

4 0
3 years ago
What is the acceleration of a vehicle that goes from 35m/s to a stop in 35 s?
nasty-shy [4]
35m/s is meters per second.
35x35=1225
it will have gone a total of 1225 meters on thirty five seconds
5 0
3 years ago
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