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2 years ago

12

PLS HELPPP!! ITS DUE TODAY!

1 answer:

Andrews [41]2 years ago

3 0

A is an example of physical change

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PLEASE HELP ASAP!!!

kirza4 [7]

Answer:

0.5mv^2=50, v=5, 25/2×m=50, m=50×2/25, So, the answer is 4

6 0

2 years ago

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

2 years ago

A 0.50 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict

Genrish500 [490]

Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

The object is released from rest when the spring is compressed 0.19 m.

(A) The force acting on the object is given by :

F = kx

$F=157\times 0.19\\\\F=29.83\ N$

(B) The force is simply given by :

F = ma

a is acceleration at that instant

$a=\dfrac{F}{m}\\\\a=\dfrac{29.83}{0.5}\\\\a=59.66\ m/s^2$

6 0

2 years ago

On Earth, the gravitational field strength is 10 N/kg. Calculate Ep for a 4 kg bowling ball have that is being held 2 m above th

Soloha48 [4]

Answer:

80 J

Explanation:

Ep = mgh

Ep = (4 kg) (10 m/s²) (2 m)

Ep = 80 J

4 0

2 years ago

100 point <br> good luck answering this<br> 2+2-4+8+-5+???

Darya [45]

The answer to that is 3

7 0

2 years ago