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3 years ago

What is the basic unit for volume

Physics

2 answers:

Marta_Voda [28]3 years ago

4 0

The basic measurement of volume is the liter.

Zigmanuir [339]3 years ago

3 0

Basic unit of length = [length] = meter

Basic unit of area = [length²] = meter²

Basic unit of volume = [length³] = meter³

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A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

$m_{1}=5.8 kg$ is the mass of the bowling ball

$V_{1}=4.34 m/s$ is the velocity of the bowling ball

$m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg$ is the mass of the ping-pong ball

$V_{2}$ is the velocity of the ping-pong ball

Now, the momentum $p_{1}$ of the bowling ball is:

$p_{1}=m_{1}V_{1}$ (1)

$p_{1}=(5.8 kg)(4.34 m/s)$

$p_{1}=25.172 kg m/s$ (2)

And the momentum $p_{2}$ of the ping-pong ball is:

$p_{2}=m_{2}V_{2}$ (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

$p_{1}=p_{2}$ (4)

$m_{1}V_{1}=m_{2}V_{2}$ (5)

Isolating $V_{2}$ :

$V_{2}=\frac{m_{1}V_{1}}{m_{2}}$ (6)

$V_{2}=\frac{25.172 kg m/s}{0.002214 kg}$ (7)

Finally:

$V_{2}=11369.46 m/s$

6 0

3 years ago

Are any rocks that form from another rock as a result of change in blank or blank

charle [14.2K]

The correct answer is rock cycle

3 0

4 years ago

A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou

Setler [38]

Answer:

a. wavelength of the sound, $\vartheta = 1.315\vartheta_{o}$

b. observed frequecy, $\lambda = 0.7604\lambda_{o}$

Given:

speed of sound source, $v_{s}$ = 80 m/s

speed of sound in air or vacuum, $v_{a}$ = 343 m/s

speed of sound observed, $v_{o}$ = 0 m/s

Solution:

From the relation:

v = $\vartheta \lambda$ (1)

where

v = velocity of sound

$\vartheta$ = observed frequency of sound

$\lambda$ = wavelength

(a) The wavelength of the sound between source and the listener is given by:

$\lambda = \frac{v_{a}}{\vartheta }$ (2)

(b) The observed frequency is given by:

$\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}$

$\vartheta = \frac{334}{334 - 80}\vartheta_{o}$

$\vartheta = 1.315\vartheta_{o}$ (3)

Using eqn (2) and (3):

$\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}$

$\lambda = 0.7604\lambda_{o}$

4 0

3 years ago

Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an

Artyom0805 [142]

a) For the motion of car with uniform velocity we have , $s = ut+\frac{1}{2}at^2$ , where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 $m/s^2$

Substituting

$520 = u*223\\ \\u = 2.33 m/s$

The constant velocity of car a = 2.33 m/s

b) We have $s = ut+\frac{1}{2} at^2$

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

$520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2$

Now we have v = u+at, where v is the final velocity

Substituting

v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 $m/s^2$

7 0

3 years ago

Three charges are located at a different position in a plane: q1= 10μC at →r1=(5,6)cm q2=−27μC at →r2=(−6,10)cm and q3=−12μC at

sasho [114]

Answer:

E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

E_{total} = 2,467 10⁶ N / A θ = -21.8

Explanation:

For this exercise we will use that the electric field is a vector quantity, so the total field is

E_total = E₁₃ + E₂₃

bold font vectors . We can work with the components of the electric field in each axis

X- axis

E_ total x = E₁₃ₓ + E_{23x}

y-axis

E_{total y} = E_{13y} + E_{23y}

the expression for the electric field is

E = k q / r²

where r is the distance between the charge and the positive test charge

in this exercise

Let's find the field created by charge 1

q₁ = 10 μC = 10 10⁻⁶ C

x₁ = 5 cm = 0.05 m

x₃ = 21 cm = 0.21 m

E_{13x} = 9 10⁹ 10 10⁻⁶ / (0.21 -0.05)²

E_{13x} = 3.516 10⁶ N / C

y₁ = 6 cm = 0.06 cm

y₃ = -12 cm = -0.12 m

E_{13y} = 9 10⁹ 10 10⁻⁶ / (-0.12 - 0.06)²

E_{13y} = 2,777 10⁶ N / C

let's find the field produced by charge 2

q₂ = -27 μC = - 27 10⁻⁶ C

x₂ = -6 cm = -0.06 m

x₃ = 0.21 m

E_{23x} = 9 10⁹ 27 10⁻⁶ / (0.21 + 0.06)²

E_{23x} = 1.23 10⁶ N / A

y₂ = 10 cm = 0.10 m

y₃ = -0.12 m

E_{23y} = 9 10⁹ 27 10⁻⁶ / (-0.12 - 0.10)²

E_{23y} = 1.86 10⁶ N / C

Taking the components we can calculate the total electric field, we must use that charge of the same sign repel and attract the opposite sign, remember that the test charge is always considered positive.

E_{total x} = E_{13x} - E_{23x}

E_{total x} = (3.516 - 1.23) 10⁶

E_{total x} = 2.29 10⁶ N / A

E_{total y} = -E_{13y} + E_{23y}

E_{total y} = (-2.777 +1.86) 10⁶ N / A

E_{total y} = -0.917 10⁶ N / A

we can give the result in two ways

E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

or in the form of modulus and angle, let's use the Pythagorean theorem to find the modulus

E_{total} = √ (E_{total x}^2 + E_{total y}^ 2)

E_{total} = √ (2.29² + 0.917²) 10⁶

E_{total} = 2,467 10⁶ N / A

let's use trigonometry for the angle

tan θ = E_total and / E_totalx

θ = tan⁻¹ E_{total y} / E_{total x}

θ = tan⁻¹ (-0.917 / 2.29)

θ = -21.8

The negative sign indicates that the angle is measured with respect to the x-axis in a clockwise direction.

7 0

3 years ago