Numatoms is initially 7. what is numatoms after: numatoms += 5?

A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t

Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

P.E = mgh joule

Considering h = 0 for the vertical position

The h at ∅ = 42° is h = L (1 - cos∅)

P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

P.E = 25 x 9.8 x 2.2 (1 - cos42°)

= 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

∴ K.E = P.E

1/2 mv² = 138.44

1/2 x 25 x v² 138.44

v² = 11.0752

v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

Tension on string, T = Force acting on the swing, F

$W=L\int\limits^0_\phi{F} \, d \phi$

$=L\int\limits^0_\phi{mg.sin \phi} \, d \phi$

= $-Lmg[cos\phi]_{42}^{0}$

= - 2.2 x 25 x 9.8 [cos0 - cos 42°]

= - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0

3 years ago

Uploaded imageTwo long, parallel wires each carry the same current I in the same direction (see figure below). The total magneti

LekaFEV [45]

Answer:

Zero

Explanation:

Two long parallel wires each carry the same current I in the same direction. The magnetic field in wire 1 is given by :

$B_1=\dfrac{\mu_oI_1}{2\pi r}$

Magnetic force acting in wire 2 due to 1 is given by :

$F_{2}=I_2lB_1$

$F_2=\dfrac{\mu_oI_1I_2 l}{2\pi r}$

Similarly, force acting in wire 1 is given by :

$F_1=\dfrac{\mu_oI_1I_2 l}{2\pi r}$ According to third law of motion, the force acting in wire 1 will be in opposite direction to wire 2 as :

$F_2=-F_1$

So, the total magnetic field at the point P midway between the wires is in what direction will be zero as the the direction of forces are in opposite direction.

7 0

3 years ago

A student rolls these four items on the same surface and at the same speed. Which item will have the greatest kinetic energy?

dimulka [17.4K]

Basketball is the answer

7 0

3 years ago

For a block to move down an inclined plane what force has to be the greatest?

Hatshy [7]

Answer:

D) True. This is what creates the body weight

Explanation:

Let's write Newton's second law for this case. For inclined planes the reference system takes one axis parallel to the plane (x axis) and the other perpendicular to the plane (y axis)

X axis

Wx -fr = ma

Y Axis

N - Wy = 0

With trigonometry we can find the components of weight

sin θ = Wₓ / W

cos θ = $W_{y}$ / W

Wₓ = W sin θ

$W_{y}$ = W cos θ

W sin θ - fr = ma

From this expression as it indicates that the body is descending the force greater is the gravity that create the weight of the body

Let's examine the answers

A False This force does not apply because it is not a spring

B) False. It is balanced at all times with the component (Wy) of the weight

C) False. For there to be a rope, if it exists you should be less than the weight component for the block to lower

D) True. This is what creates the body weight

E) False. The cutting force occurs for force applied at a single point and gravity is applied at all points

5 0

3 years ago

A 500 G ball and a 200 g or dropped from a tower. Those balls are the same size and there is no air resistance. Which one hits t

ExtremeBDS [4]

They both hit at the same time

6 0

3 years ago

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