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3 years ago

How far will an object move in 6 s if its average

Physics

1 answer:

Dafna11 [192]3 years ago

6 0

Given:-

Time taken by the particle (t) = 6 s
Average speed (v) = 40 m/s

To Find: Distance (s) travelled by the particle.

We know,

s = vt

where,

s = Distance travelled,
v = Speed &
t = Time taken.

Putting the values,

s = (40 m/s)(6 s)

→ s = 240 m ...(Ans.)

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Simplify 45÷3+2×8-12+42

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Answer:

Explanation is in the picture and the answer is 16

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3 years ago

Two sound waves, from two different sources with the same frequency of 420 Hz travel in the same direction at 336 m/s. The sourc

Tresset [83]

Answer:

Pi(3.14) radians or 180º degrees

Explanation:

First of all, we need to obtain the wavelength of a wave traveling to the speed of sound and 420 Hz of frequency.

The formula is:

$l=wavevelocity/frequency$

where l = wavelength in meters

With current values:

l = 336 [m/s]/420[1/s] = 0.8 meters

Since a complete cycle (360º or 2pi radians) needs 0.8 meters to complete, 0.4 meters or 40 cm is just half of it, making a 180º degree phase or 3.14 radians.

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3 years ago

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3 years ago

Read 2 more answers

In Scientific Notation, numbers are written in the form a x 10b . In this form, b is called the

ivann1987 [24]

Answer:

b is called the power of base 10.

Explanation:

In the scientific notation form

$a\,\,10^b$

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3 years ago

Add a vector whose magnitude is 13 with angle 27 degrees to one whose magnitude is 11 with angle 45 degrees? Put the length firs

mafiozo [28]

Answer:

Magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

Explanation:

Magnitude of first vector = $|A| = 13\ \text{units}$

Angle = $\theta_1=27^{\circ}$

Magnitude of second vector = $|B| = 11\ \text{units}$

Angle = $\theta_2=45^{\circ}$

x component of first vector

$A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}$

y component of first vector

$A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}$

x component of second vector

$B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}$

y component of first vector

$B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}$

Adding the magnitudes

$C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}$

$C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}$

Magnitude of the sum of the vectors would be

$|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}$

The direction would be

$\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}$

The magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

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3 years ago