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Natasha_Volkova [10]
2 years ago
6

Jonathan and Cody’s older brother Josh, who is pictured in the Figure below, is standing at the top of a half-pipe at Newton’s S

kate Park. Gravity is exerting a downward force on the skateboard as seen in the picture. Why doesn’t it tip over the edge and start rolling down the side of the half-pipe?
Physics
1 answer:
Margarita [4]2 years ago
8 0
In order to begin moving a Mass(Accelaration) will solve for the force required to move the skateboard
Mass(9.8) will give you the force needed to move the skateboard originally
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Martin has hypothesized that the size and types of trees differ between valleys and higher elevations. He sets up a transect lin
pogonyaev

Answer: The dependent Variables include: Size and Types of trees.

The independent Variables include:

geographic elevations that is, valleys and the higher elevations.

Explanation:

The dependent variable simply refers to the variable a researcher tests or measures during an experiment. On the other hand, the independent variable simply refers to the variable that's controlled during an experiment.

Based on the definition above, the dependent variables include the size and the types of trees while the independent variables include the

geographic elevations that is, the valleys and the higher elevations.

3 0
3 years ago
On a frictionless track,cart 1 is moving with a constant,rightward(+) velocity of 1.0m/s.Cart 2 is also moving rightward with co
mamaluj [8]

Answer:

The final velocity of cart 1 is 3m/s

Explanation:

From principle of conservation of linear momentum, which states that sum of the momentum before collision is equal to the sum of the momentum after collision.

Momentum, P is given as mass x velocity.

ΔP = Δmv = m₁u₁ +m₂u₂ = m₁v₁ + m₂v₂

Assumptions:

  • If the two carts are moving on frictionless track, then limiting frictional forces due to their weights are negligible.
  • After the elastic collision, the two carts will move separately with different velocity

u₁ + u₂ = v₁ + v₂;

where;

u₁ and u₂  are the initial velocity for cart 1 and cart 2 respectively

v₁ and v₂  are the final velocity for cart 1 and cart 2 respectively

1 m/s + 5 m/s = v₁  + 3m/s

6 m/s =  v₁  + 3m/s

v₁  = 6 m/s - 3m/s = 3m/s

Therefore, the final velocity of cart 1 is 3m/s

4 0
3 years ago
EJ is taking his little sister, Ebony, on her first carousel ride. His mass is 75.0 kg and her mass is 10.0 kg. They ride the sa
Bad White [126]
<span>V = 11*1.363 = 15m/s = 33.6mi/hr </span>
3 0
3 years ago
Read 2 more answers
Pulling a wheel with a string? A bicycle wheel is mounted on a fixed, frictionless axle, with a light string wound around its ri
m_a_m_a [10]
A)♣ the string being pulled, 
<span>the angular speed is: w1=w0 +w’*t, hence t=(w1-w0)/w’; </span>
<span>the angular path is: b=w0*t+0.5*w’*t^2, where angular acceleration </span>
<span>w’=T/J, torq T=F*r, and b*r=L; </span>
<span>♦ thus b=w0*(w1-w0)/w’ +0.5*w’*((w1-w0)/w’)^2 = </span>
<span>= (w1-w0)*(w0 +0.5w1 -0.5w0)/w’ =0.5*(w1^2 –w0^2)/w’; </span>
<span>♠ and b=L/r =0.5*(w1^2 –w0^2)/(F*r/J); </span>
<span>2(F*r/J)*L/r =w1^2 –w0^2, hence </span>
<span>w1^2=2F*L/J +w0^2; </span>
<span>b)♣ the power is P=F*(w0*r);

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>
6 0
3 years ago
The mass of Tesla Model S is 2000 kg and its engine produces
yulyashka [42]

Answer:

The ratio of the power generated by the engines are \dfrac{2}{3}

(b) is correct option.

Explanation:

Given that,

Mass of Tesla = 2000 Kg

Torque = 400 Nm

Mass of Mercedes = 1.5 M Tesla

The Benz engine produces 50% more torque than the Tesla.

We need to calculate the power generated by the engines 6 seconds after a start from rest

Using formula of  power

P=F\cdot v

We know that,

v=at

v=\dfrac{F}{m}t

v=\dfrac{\tau}{mr}t

Put the value of F and v in equation (I)

P=\dfrac{\tau}{r}\times\dfrac{\tau}{mr}t

P=\dfrac{\tau^2t}{mr^2}

Here, r and t are same for both bodies.

P\propto\dfrac{\tau^2}{m}

We need to calculate the ratio of the power generated by the engines

Using formula of power

\dfrac{P_{T}}{P_{B}}=\dfrac{\tau_{T}^2}{m_{T}}\times\dfrac{m_{B}}{\tau_{B}}

Put the value into the formula

\dfrac{P_{T}}{P_{B}}=\dfrac{400^2}{2000}\times\dfrac{3000}{600^2}

\dfrac{P_{T}}{P_{B}}=\dfrac{2}{3}

Hence, The ratio of the power generated by the engines are \dfrac{2}{3}

3 0
3 years ago
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