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kumpel [21]
3 years ago
7

if an object is moving with a velocity of 24m/s and has an acceleration of -4m/s how long will it take to stop

Physics
1 answer:
ololo11 [35]3 years ago
7 0
The negative sign on the acceleration is only a vector quantity that means the object is accelerating to the left. Hence, we can only focus on it magnitude which is 4 m/s^2. Acceleration is the change in velocity over time. The change in velocity must be 24 m/s - 0 m/s, if you want the object to stop. Therefore,

a = (v2 - v1)/t
4 = (24 - 0)t
t = 6 seconds

The object will stop after 6 seconds.
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Answer:

magnatic force can be created

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What is the mass of a bird if it has a potential energy of 147J when flying at a height of 30 m above the ground?
Artist 52 [7]

Answer:

0.5kg

Explanation:

Given parameters:

Potential energy = 147J

Height  = 30m

Unknown:

Mass of the bird  = ?

Solution:

Potential energy is the energy due to the position of a body. Now, the expression for finding the potential energy is given as;

      P.E  = mgH

m is the mass

g is the acceleration due to gravity = 9.8m/s²

H is the height

      147  = m x 9.8 x 30

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5 0
3 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

         

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HACTEHA [7]

On behalf of the state, write an invitation for the siblings for the gala dinner


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3 years ago
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oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

d=\frac{1,280,000,000km}{37,000}=34,594.59km

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}

hence, the speed of the Hubble is approximately 384km/min

5 0
3 years ago
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