I throw a ball upward at an initial speed of 20 m/s. How much time does it take before the ball slows to a speed of 0 m/s

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

3 years ago

Un soldado de 1.80 m de altura ha sido herido por una bala, cuya masa es de 100 g. Deciden colocar al soldado en una centrifugad

maks197457 [2]

eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

3 0

3 years ago

In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables

Juli2301 [7.4K]

They made me do it I don’t even know what to say I’m so sorry

7 0

3 years ago

A 2.0 m tall person is standing in front of a flat vertical mirror. What minimum height must the mirror have if the person is to

dangina [55]

Answer:

100cm

Explanation:

Since the eyes are 6 cm below the top of her head, the point of incidence of the ray must be

200cm-3cm=197cm

Since the eyes are 194 cm from her feet, the point of incidence of this ray must be

194cm/2=97cm

So the lower edge of the mirror must be 97 cm from the ground and the vertical dimension of the mirror must be 197 cm - 97 cm = 100 cm, which is half the height of the person.

7 0

3 years ago

A 40 kg skier starts at the top of a 12-meter high slope at the bottom she is traveling 10 m/s how much energy does she lose fri

Iteru [2.4K]

Potential energy at top:
PE = mgh
PE = 40 x 9.81 x 12
P.E = 4,708.8 J

Kinetic energy at bottom:
KE = 1/2 mv²
KE = 1/2 x 40 x 10²
K.E = 2,000 J

P.E = K.E + Frictional losses
Frictional losses = 4708 - 2000
Frictional losses = 2708 J

The answer is D.

7 0

3 years ago

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