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3 years ago

If I move 15ft foward, 15 ft backwards, 15 ft to the right, 15ft to the left where am I?

Physics

2 answers:

Naya [18.7K]3 years ago

8 0

Exactly where you started

lisov135 [29]3 years ago

4 0

-- When you finish those moves, you are exactly where you started from.

-- Your distance covered is 60 ft.

-- Your displacement is zero.

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Sometimes may cause involuntary responses like twitching

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3 years ago

Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1

Nonamiya [84]

Answer:

$a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

Magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$

Explanation:

$t_1=2\ \text{s}$

$v_x=1\ \text{m/s}$

$v_y=3\ \text{m/s}$

$t_2=2.5\ \text{s}$

$v_x=4\ \text{m/s}$

$v_y=3\ \text{m/s}$

Average acceleration in the different axes

$a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2$

$a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2$

The components of the acceleration is $a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

The magnitude of acceleration

$a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2$

Direction

$\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}$

The magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$ .

7 0

2 years ago

Two coils have the same number of circular turns and carry the same current. Each rotates in a magnetic field acting perpendicul

ANEK [815]

Answer:

4.14 cm.

Explanation:

Given,

For Coil 1

radius of coil, r₁ = 5.6 cm

Magnetic field, B₁ = 0.24 T

For Coil 2

radius of coil, r₂ = ?

Magnetic field, B₂ = 0.44 T

Using formula of maximum torque

$\tau_{max}= NIAB$

Since both the coil experience same maximum torques

now,

$NIA_1B_1 = NIA_2B_2$

$A_1B_1 = A_2B_2$

$r_1^2 B_1 = r_2^2 B_2$

$5.6^2\times 0.24= r_2^2\times 0.44$

$r_2 = 4.14\ cm$

Radius of the coil 2 is equal to 4.14 cm.

4 0

3 years ago