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3 years ago

What is the relationshop between a pure substance and a mixture

1 answer:

7 0

The difference between a pure substance and a mixture is that mixtures are combinations of two or more pure substances, while pure substances cannot be separated into any other kinds of matter.

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Cart A and cart B are traveling in the same direction on a straight track. Cart A is moving at 9.25 m/s and cart B is moving at

Liula [17]

Answer:

The speed of cart B is 11.21 m/s.

Explanation:

Given that,

Speed of cart A = 9.25 m/s

Speed of cart B = 7.15 m/s

Mass of cart A = 72.0 kg

Mass of cart B = 55.0 kg

Speed of card A after collision = 6.15 m/s

We need to calculate the speed of cart B

Using conservation of momentum

$m_{A}v_{A}+m_{B}v_{B}=m_{A}v_{A}+m_{B}v_{B}$

Put the value into the formula

$72.0\times9.25+55.0\times7.15=72.0\times6.15+55.0\times v_{B}$

$v_{B}=\dfrac{72.0\times9.25+55.0\times7.15-72.0\times6.15}{55.0}$

$v_{B}=11.21\ m/s$

Hence, The speed of cart B is 11.21 m/s.

7 0

3 years ago

Charges of 4.0 μC and −6.0 μC are placed at two corners of an equilateral triangle with sides of 0.10 m. What is the magnitude o

jek_recluse [69]

Answer:

4.763 × 10⁶ N/C

Explanation:

Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.

Resolving E₂ into horizontal and vertical components, we have

E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.

Summing the horizontal components we have

E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²

= -k/r²(q₁ + q₂cos60)

= -k/r²(4 μC + (-6.0 μC)(1/2))

= -k/r²(4 μC - 3.0 μC)

= -k/r²(1 μC)

= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²

= -9 × 10⁵ N/C

Summing the vertical components, we have

E₄ = 0 + (-E₂sin60)

= -E₂sin60

= -kq₂sin60/r²

= -k(-6.0 μC)(0.8660)/(0.10 m)²

= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²

= 46.77 × 10⁵ N/C

The magnitude of the resultant electric field, E is thus

E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴

= 476.28 × 10⁴ N/C

= 4.7628 × 10⁶ N/C

≅ 4.763 × 10⁶ N/C

8 0

3 years ago

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0

3 years ago

What is the momentum of an object that is moving in a straight line at a speed of 10 m/s and has a mass of 10 kilograms?

Aleksandr-060686 [28]

The momentum of an object is given by the product between its mass and its velocity:
$p=mv$
where m is the mass and v the velocity.

For the object in our problem, m=10 kg and v=10 m/s, therefore its momentum is
$p=mv=(10 kg)(10 m/s)=100 kg m/s$
So, the correct answer is B).

8 0

3 years ago

A 15.0-kg child descends a slide 2.40 m high and reaches the bottom with a speed of 1.10 m/s .

pickupchik [31]

The thermal energy that is generated due to friction is 344J.

<h3>What is the thermal energy?</h3>

Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.

Thus, the kinetic energy is given as;

KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J

PE = mgh = 15.0-kg * 9.8 m/s^2 * 2.40 m = 352.8 J

The thermal energy is; 352.8 J - 9.1 J = 344J

Learn more about thermal energy due to friction:brainly.com/question/7207509

#SPJ1

7 0

2 years ago