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3 years ago

What is the vector sum of 5km east and 5km west

Physics

2 answers:

zvonat [6]3 years ago

3 0

If you travel 5km east then 5km west you end up back where you started therefore the vector sum would be 0.

NeX [460]3 years ago

3 0

Answer:

Zero

Explanation:

Let us consider the east direction to be positive then west is considered to be negative.

Net displacement = 5 km - 5 km

= 0 km

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A flute is designed so that it plays a frequency of 261.6 Hz, middle C, when all the holes are covered and the temperature is 20

irinina [24]

Answer:

0.655 m

13.468°C

Explanation:

v = Speed of sound at 20.0°C = 343 m/s (general value)

For one both end open we have the expression

$L=\dfrac{\lambda_1}{2}\\\Rightarrow L=\dfrac{\dfrac{v}{f_1}}{2}\\\Rightarrow L=\dfrac{\dfrac{343}{261.6}}{2}\\\Rightarrow L=0.655581039755\ m$

The length of the flute is 0.655 m

Beat frequency is given by

$\Delta f=f_1-f_2\\\Rightarrow 3=261.6-f_2\\\Rightarrow f_2=261.6-3\\\Rightarrow f_2=258.6\ Hz$

Velocity of the wave is

$v=f_2\lambda_1\\\Rightarrow v=258.6\times \dfrac{343}{261.6}\\\Rightarrow v=339.066513761\ m/s$

The temperature is given by

$T=273(\dfrac{v}{331})^2\\\Rightarrow T=273(\dfrac{339.066513761}{331})^2\\\Rightarrow T=286.468227799\ K=286.468227799-273=13.468227799^{\circ}C$

The temperature of the room is 13.468°C

4 0

3 years ago

Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.

Fofino [41]

Answer:

Part a)

$\alpha = 782.6 rad/s^2$

Part B)

$\omega = 587 rad/s$

Part c)

$a_t = 24.3 m/s^2$

Explanation:

Part a)

As we know that

$a = R \alpha$

so we will have

$a = 1.80 m/s^2$

$R = 0.230 cm$

$\alpha = \frac{a}{R}$

$\alpha = \frac{1.80}{0.230 \times 10^{-2}}$

$\alpha = 782.6 rad/s^2$

Part B)

Angular speed of the yo-yo

$\omega = \alpha t$

so we have

$\omega = 782.6 \times 0.750$

$\omega = 587 rad/s$

Part c)

Tangential acceleration is given as

$a_t = R \alpha$

$a_t = (3.10 \times 10^{-2})(782.6)$

$a_t = 24.3 m/s^2$

6 0

4 years ago

In the diagram, R1 = 40.0 ,

Nostrana [21]

Answer:

51 Ω.

Explanation:

We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:

Resistor 1 (R₁) = 40 Ω

Resistor 3 (R₃) = 70.8 Ω

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?

Since the two resistors are in parallel connection, their equivalent can be obtained as follow:

R₁ₙ₃ = R₁ × R₃ / R₁ + R₃

R₁ₙ₃ = 40 × 70.8 / 40 + 70.8

R₁ₙ₃ = 2832 / 110.8

R₁ₙ₃ = 25.6 Ω

Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω

Resistor 2 (R₂) = 25.4 Ω

Equivalent Resistance (Rₑq) =?

Rₑq = R₁ₙ₃ + R₂ (series connection)

Rₑq = 25.6 + 25.4

Rₑq = 51 Ω

Therefore, the equivalent resistance of the group is 51 Ω.

4 0

3 years ago

Which of the following describes the reactants of a chemical reaction

AlekseyPX

The options attached to the question are given below:

A. The substances that are formed.

B. The substances that are changed.

C. The starting materials.

D. The chemical ingredients

ANSWER

The correct option is C.

A chemical reaction is made up of two distinct parts, which are reactants and products. The reactants refers to the starting material of the chemical process, which react together under suitable conditions. The products on the other hand refers to the new substance that is formed as a result of the reaction of the reactants. The reactants are usually find at the left side of chemical equations while the product is found at the right.

7 0

4 years ago

n the railroad accident, a boxcar weighting 200 kN and traveling at 3 m/s on horizontal track slams into a stationary caboose we

USPshnik [31]

Answer:

ΔK = -6 10⁴ J

Explanation:

This is a crash problem, let's start by defining a system formed by the two trucks, so that the forces during the crash have been internal and the moment is preserved

initial instant. Before the crash

p₀ = m v₁ + M 0

final instant. Right after the crash

p_f = (m + M) v

p₀ = p_f

mv₁ = (m + M) v

v = $\frac{m}{m+M} \ v_1$

we substitute

v = $\frac{20}{20+40}$ 3

v = 1.0 m / s

having the initial and final velocities, let's find the kinetic energy

K₀ = ½ m v₁² + 0

K₀ = ½ 20 10³ 3²

K₀ = 9 10⁴ J

K_f = ½ (m + M) v²

K_f = ½ (20 +40) 10³ 1²

K_f = 3 10⁴ J

the change in energy is

ΔK = K_f - K₀

ΔK = (3 - 9) 10⁴

ΔK = -6 10⁴ J

The negative sign indicates that the energy is ranked in another type of energy

7 0

3 years ago