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bezimeni [28]
3 years ago
8

Rohith has been jogging to the bus stop for 2.0 minutes at 3.5 m/s when he looks at

Physics
1 answer:
nikdorinn [45]3 years ago
6 0
Multiply these numbers and there’s your answer
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The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t
coldgirl [10]

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

    Y axis

     N- W = 0

     N = W = mg

  X axis

     -Fr = m a

     -μ N = m a

     -μ mg = ma

     a = μ g

     a  = - 0.32 9.8

     a =  - 3.14 m/s²

We calculate the distance using the kinematics equations

    Vf² = Vo² + 2 a x

     x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

 Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

     x = ( 0 - 15²) / 2 (-3.14)

     x=  35.8 m

The shortest braking distance is  35.8 m

7 0
3 years ago
How are thunderstorms kept alive?
Burka [1]

Answer:

E step leader

Explanation:

5 0
3 years ago
Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

4 0
3 years ago
How would convection warm up an iguana cage
harina [27]
Using a basking spot so some sort of heated object, for example heating lamp or heating pad.
3 0
3 years ago
When the distance between two objects doubles, what happens to the gravitational attraction between these two objects?
Feliz [49]

Answer:

As indicated by Newton's law of attraction each article or body in the universe draws in every single item towards one another and that power of fascination is straightforwardly relative to the result of their masses and contrarily corresponding to the square of the distance between them.  

The power of gravity between two articles will diminish as the distance between them increments. The two most significant elements influencing the gravitational power between two items are their mass and the distance between their focuses. As mass increments, so does the power of gravity, however an increment in distance mirrors a reverse proportionality, which makes that power decline dramatically.  

At that point by Newton's All inclusive Law of Attractive energy;  

F=GMm/R^2  

Mm= result of the majority  

R=Distance Between the two masses by focus.  

On the off chance that R is multiplied, new force=GMm/(2R)^2  

=GMm/4R^2  

Unique Power/New Force=4/1  

F/4=New Power

8 0
3 years ago
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