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Bogdan [553]
2 years ago
7

A 720 kg roller-coaster starts off from Location A. Assuming friction does not impede the car's motion, what will be the change

in its potential energy by moving from Location A to Location B
Physics
2 answers:
Alekssandra [29.7K]2 years ago
8 0

The correct answer is: 6.0 x 10^5 J

vaieri [72.5K]2 years ago
7 0
We know, Potential Energy = m * g * h
Here, mass & gravity would be same, but their height will change so it will be:

ΔU = U₂ - U₁
ΔU = mgh₂ - mgh₁
ΔU = mg (h₂ - h₁)

Hope this helps!
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Blocks A and B of unknown masses m1 and m2, respectively, are set up on an inclined plane as shown. Block A is attached to block
Korvikt [17]

Newton's second law we can find that the correct answer is:  

 E)  It cannot be determiner whick block has more masses from the information provided

Newton's second law establishes the relationship between force, mass, and acceleration of a body. Since force and acceleration are vector quantities, their components must be added on each axis

For this problem we have two bodies, let's write Newton's second law for the body B, we assume that the body B descends

            W_b - T = m_b a

            W_b  = m_b g

            m_b - T = m_b a

Where W_b is the weight of block B, T the tension of the string, mb the mass of block b and the acceleration

Now let's find the relation for block A

let's set a datum with the x axis parallel to the ramp

           T - Wₓ = mₐ a

           sin θ = Wₓ / W

            Wₓ = Wₐ sin θ

             Wₐ = mₐ g

Where Wₓ is the component of the weight, Wₐ the weight of the body A and θ the angle of the plane

Let's write our system of equations

           m_b g - T = m_b a

           T - mₐ g sin θ = mₐ a

let's add the equations

            g (m_b - mₐ sin θ) = (m_b + mₐ) a

            a =   \frac{m_b - m_a \ sin  \ \theta}{m_b+m_a} \ g

Let's analyze this expression

  • The numerator is positive the body B descends, this occurs when

          m_b - mₐ sin θ > 0

           

  • The numerator is negative, body B rises

           m_b - mₐ  sin θ <0

We can observe that the acceleration is positive or negative depending on the relation of the masses and the angle of the plane.

In conclusion using Newton's second law we find that the correct answer is  

 E )   It cannot be determiner whick block has more masses from the information provided

learn more about Newton's second law here:

brainly.com/question/9099891

8 0
2 years ago
Robbie found a rock in a stream. The rock was smooth and round. What MOST LIKELY caused this to happen?
Degger [83]

Answer:

C, weathering by the water.

Explanation:

While in the river, it scraps againsts other rocks and things, which causes it to change shape. For example be smoother and round.

7 0
2 years ago
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

F_D = c_D A \frac{\rho V^2}{2}

Where,

F_D = Drag Force

c_D = Drag coefficient

A = Area

\rho= Density

V = Velocity

Our values are given by,

c_D = 0.5 (That is proper of a cone-shape)

A = 9m^2

\rho = 1.2Kg/m^3

V = 6.5m/s

Part A ) Replacing our values,

F_D = 0.5*9*\frac{1.2*6.5^2}{2}

F_D = 114.075N

Part B ) To find the torque we apply the equation as follow,

\tau = F*d

\tau = (114.075N)(7)

\tau = 798.525N.m

3 0
3 years ago
As a pendulum bob swings back and forth several times, the maximum height it reaches becomes less and less.
Ymorist [56]

Answer:

A or B

Explanation:

4 0
3 years ago
Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109
r-ruslan [8.4K]

Answer:

change in mass = 2.41*10^{8}kg

Explanation:

The change in the mass can be computed by using the relation

E=\Delta mc^2\\\Delta m=\frac{E}{c^2}(1)

That is, the energy liberated comes from the mass of the nuclear fuel. The energy generated in one year is

E=Pt=2.3*10^{9}\frac{J}{s}*1 year*\frac{365.25 day}{1 year}*\frac{24h}{1 day}*\frac{3600s}{1h}=7.25*10^{16}J

Hence, by replacing in the equation (1) you have  (c=3*10^{8}m/s)

\Delta m=\frac{7.25*10^{16}J}{3*10^{8}\frac{m}{s}}=2.41*10^{8}kg

HOPE THIS HELPS!!

3 0
3 years ago
Read 2 more answers
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