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3 years ago

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According to Charles’s law, when the temperature of a gas increases at constant pressure its a. volume increases. b. mass increa

ses. c. volume decreases. d. particles move more slowly.

1 answer:

xxMikexx [17]3 years ago

8 0

The answer is A, the volume increases.
hope this helps

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Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s

enot [183]

Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:

$p_i = p_f$

$m_1u_1 + m_2v_2 = (m_1 + m_2)v$

$v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}$

$v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s$

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}$

Here, initial velocity is the final velocity from the first stage. Therefore:

$v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s$

3. Use conservation of momentum to find the combined speed of Gayle and her brother.

Given:

Initial velocity of Gayle and sled is, $u_1(i)=10.5$ m/s

Initial velocity of her brother is, $u_2(i)=0$ m/s

Mass of Gayle and sled is, $m_1=55.0$ kg

Mass of her brother is, $m_2=30.0$ kg

Final combined velocity is given as:

$v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}$

$v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79$ m/s

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s$

6 0

3 years ago

Car A has a mass of 1,200 kg and is traveling at a rate of 22km/hr. It collides with car B has a mass of 1,900 kg and is traveli

Zina [86]

<span>The momentum before the collision is equal to the momentum after the collision</span>

7 0

3 years ago

Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag

Salsk061 [2.6K]

Answer:

The resultant electric force is 14.8N to the right.

Explanation:

Since the three charges aren't in the same line, we have to break down the force in components. First, we need to know the distance from the third charge to the other ones. That is made using the Pythagorean Theorem. As the figure is symmetric with respect to the x-axis, the two distances are the same:

$r=\sqrt{(0.50m)^{2}+(0.70m)^{2}}=0.86m$

Now, we use the Coulomb's Law to obtain the magnitude of the individual forces caused by each charge on the third charge:

$|F_{13}|=k\frac{q_1q_3}{r^{2}} \\\\|F_{13}|=(9*10^{9}Nm^{2}/C^{2})\frac{(2.5*10^{-9}C)(3.0*10^{-9}C)}{(0.86m)^{2}}\\\\|F_{13}|=9.1N$

For the same reason the distances are the same, the magnitude of the forces are the same:

$|F_{23}|=|F_{13}|=9.1N$

So, to get the resultant force, we have to break down this forces in components. To do this, we need their angles with respect to the x-axis. Let θ₁ and θ₂ be these angles, respectively. Then, we calculate them using trigonometry:

$\theta_1=\arctan(\frac{-0.50m}{0.70m})=-35.5\°\\\\\theta_2=\arctan(\frac{0.50m}{0.70m})=35.5\°$

Now, we calculate the components of the forces:

$F_{13}_x=F_{13}\cos\theta_1=9.1N\cos(-35.5\°)=7.4N\\\\F_{13}_y=F_{13}\sin\theta_1=9.1N\sin(-35.5\°)=-5.3N\\\\F_{23}_x=F_{23}\cos\theta_2=9.1N\cos(35.5\°)=7.4N\\\\F_{23}_y=F_{23}\sin\theta_2=9.1N\sin(35.5\°)=5.3N$

Evidently, the y-components cancel out, and the resultant electric force on the third charge is $7.4N+7.4N=14.8N$ along the x-axis (to the right, because it's positive).

8 0

3 years ago

What acceleration of a 45 kg mass pushed with 15 N of force?

Hitman42 [59]

Explanation:

Fnet = ma.

Assume that there is no resistance acting on the mass. Therefore a = 15N / 45kg = 0.33N/kg or 0.33m/s^2.

3 0

3 years ago

Four identical metallic objects carry the following charges 1.82 6.65 4.80 and 9.30 C The objects are brought simultaneously int

GuDViN [60]

Answer:

a) 5.64 C

b) 3.5*10¹⁹ protons

Explanation:

a)

Since the four metallic objects are identical, and total charge must be conserved, this means that after brought simultaneously into contact so that each touches the others, once separated, total charge must be the same than before being brought in contact.
But due they are identical, after charges were able to transfer freely between them, the four objects must have the same final charge, i.e. the fourth part of the total charge, as follows:

$Q_{n} = \frac{Q_{tot}}{4} = \frac{22.57C}{4} = 5.64 C (1)$

b)

This charge will be divided between n protons, since the charge is positive.
Since each proton carries a charge equal to the elementary charge e, which value is 1.6*10⁻¹⁹ C, we can find the number of protons in excess, doing the following calculation:
$n_{p} =\frac{Q_{n}}{e} = \frac{5.64C}{1.6e-19C} = 3.5 e19 C (2)$

7 0

3 years ago