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anyanavicka [17]

3 years ago

12

An airplane travels directly from Washington, D.C., to Atlanta, Georgia, a distance of 850 km at a velocity of 425 km/h southwes

t
How long does the trip take in hours?

1.0 h

2.0 h

0.5 h

4.0 h

1 answer:

Sloan [31]3 years ago

4 0

Answer:

obviously 2 hours cuz 850/425

Explanation:

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A solid is stirred into a liquid and dissolves. Which type of mixture forms?

AVprozaik [17]

The type of mixture that is formed when a solid is stirred into a liquid and dissolves is called suspension. The particles involved or being mixed in this type of mixture is large enough that can be seen by the naked eye without the aid of any device. A suspension mixture has a heterogeneous mixture.

5 0

3 years ago

Read 2 more answers

A 6 kg cart starting from rest rolls down a hill and at the bottom has a speed of 10 m/s. What is the height of the hill?

Arisa [49]

Answer:

h = 5.09 m

Explanation:

Applying the Law of conservation of energy to this situation, we can write:

$Kinetic\ Energy\ Gained\ by\ the\ Cart = Potential\ Energy\ Lost\ by\ the\ Cart\\\frac{1}{2}mv^2 = mgh\\\\h = \frac{v^2}{2g}$

where,

h = height of the hill = ?

v = speed of cart at the end = 10 m/s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$h = \frac{(10\ m/s)^2}{(2)(9.81\ m/s^2)}\\\\$

<u>h = 5.09 m</u>

4 0

3 years ago

Tom is throwing an baseball at an aluminum can,

pishuonlain [190]

Answer:

The question relates to the conservation of energy principle, the conservation of the linear momentum, and Newton's Laws of motion

Part A

1) Tom throwing a baseball at a can

The initial velocity of the baseball = v₂

The initial kinetic energy of the baseball, K.E.₂ = (1/2)·m₂·v₂²

∴ The final kinetic energy of the baseball, K.E.₂' = (1/2)·m₂·v₂'² < (1/2)·m₂·v₂²

Therefore, the energy of the ball before the collision is lesser than the energy of the ball after the collision

2) The evidence that would likely support the claim is that the baseball's height above the ground reduces rapidly immediately after the collision which is due to the reduced velocity, and therefore, the reduced (kinetic) energy

The final velocity of the baseball v₂' < v₂

Part B

1) The argument

The initial velocity of the can = v₁ = 0 (The can is initially at rest)

The initial kinetic energy of the can, K.E.₁ = (1/2)·m₁·v₁² = 0

The final velocity of the can v₁' > v₁ = 0

∴ The final kinetic energy of the can, K.E.₁ = (1/2)·m₁·v₁² > 0

Given that the velocity of the can increases from zero to a positive value after collision with the baseball, the kinetic energy of the can is increased from zero before the collision to a positive value after the collision

2) An evidence in support of the argument is the motion of the can which was initially at rest which is an indication of increase in energy podded by the can

Explanation:

8 0

3 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

5 0

3 years ago

Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as

viva [34]

Answer:

a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s

Explanation:

It is angular momentum given by

L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

${array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]$

Let's apply this relationship to our case

Let's start by breaking down the speed

v₀ₓ = v₀ cosn 45

voy =v₀ sin 45

v₀ₓ = 9 cos 45

voy = 9 without 45

v₀ₓ = 6.36 m / s

voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

vfy² = voy²- 2 g y

y = voy² / 2g

y = (6.36)²/2 9.8

y = 2.06 m

Let's calculate the angular momentum

L= $\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]$

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^ Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

R = vo² sin 2θ / g

R = 9² sin (2 45) /9.8

R = 8.26 m

L = $\left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]$

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^ kg m² /s

5 0

3 years ago