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Phantasy [73]
3 years ago
5

A 3.00 kg cart on a frictionless track is pulled by a string so that it accelerates at 2.00 m/s/s. what is the tension in the st

ring

Physics
2 answers:
AnnZ [28]3 years ago
5 0
Maybe the picture helps. The blue block represents the cart with a mass of 3 kg. The person(black block) is pulling the cart to the right with a force F so that the acceleration a is 2 m/s². According to Newton's 2nd law: F = m*a.

Sedaia [141]3 years ago
4 0

<u>Answer:</u> The tension in the string is 6 N

<u>Explanation:</u>

Tension is defined as the pulling force that is applied on a string, chain etc.. Force is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

F=ma

where,

F = tension or force exerted in the string

m = mass of the cart pulling the string = 3.00 kg

a = acceleration of the string = 2.00m/s^2

Putting values in above equation, we get:

F=3.00kg\times 2.00m/s^2\\\\F=6N

Hence, the tension in the string is 6 N

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White light falls on a yellow filter,if:
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Answer:

all colours are absorbed except for the colour of the filter.

Explanation:

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6 0
3 years ago
An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f
konstantin123 [22]
A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force  F=m \frac{v^2}{r}, directed toward the center of the circle (so, horizontally)
- the weight of the plane: W=mg, downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N
On the vertical axis, we have two forces: the weight
W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to a=12 m/s^2, we can find the magnitude of this other force F by using Newton's second law:
F+mg=ma
F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
R= \sqrt{(F_c^2+(W+F)^2}=
= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2}  =2.23 \cdot 10^6 N

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
\tan \theta =  \frac{-(W+F)}{F_c}= -0.5
And so, the angle is
\theta = \arctan (-0.5)=-26.8 ^{\circ}
 
7 0
3 years ago
Two balls with masses of 2.0 kg and 6.0 kg travel toward each other at speeds of 12 m/s and 4.0 m/s, respectively. If the balls
Alina [70]

Answer:

The kinetic energy lost in the collision is 48 J

Explanation:

Given;

mass of the first ball, m₁ = 2.0 kg

mass of the second ball, m₂ = 6.0 kg

initial speed of the first ball, u₁ = 12 m/s

initial speed of the second ball, u₂ = 4 m/s

let v be the final velocity of the two balls after the inelastic collision

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 12 + 6 x 4 = v(2 + 6)

48 =  8v

48 / 8 = v

v = 6 m/s

The initial kinetic energy of the balls is calculated as;

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²

K.E₁ = 144 + 48

K.E₁ = 192 J

The final kinetic of the balls is calculated as;

K.E₂ = ¹/₂(m₁ + m₂)(v²)

K.E₂ = ¹/₂(2 + 6)(6²)

K.E₂ = ¹/₂(8)(6²)

K.E₂ = 144 J

The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J

Therefore, the kinetic energy lost in the collision is 48 J

5 0
3 years ago
Which of the following statements is true?
xenn [34]

Correct answer choice is :


C) The freezing and melting temperatures of a substance are the same.


Explanation:


Fluids have a particular temperature at which they convert into solids, identified as their freezing point. In theory, the melting point of a solid should be the same as the freezing point of the liquid. In practice, small variations among these measures can be seen. The freezing point of a matter is the same as that substance's melting point. At this distinct temperature, the substance can exist as either a solid or a liquid. At temperatures below the freezing/ melting point, the substance is a solid.


4 0
3 years ago
Read 2 more answers
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