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Gala2k [10]
3 years ago
8

What happens to the acceleration when the velocity is zero

Physics
1 answer:
Ulleksa [173]3 years ago
7 0
<h2>Answer:</h2>

It's easy to fall into the temptation to say that when the velocity is zero, then the acceleration is also zero. But wait! To answer this question we need to bring out the concept of instantaneous velocity. This type of velocity stands for a specific moment, a specific instant of time, that is, t=1, \ t=2, \ t=3, \ t=3.2 \ t=4.5. If so, then acceleration may not be zero when velocity is zero. For example, suppose you throw an object upward, when it is at the top of the travel the instantaneous velocity is zero because it changes from positive to negative value and there is a moment when it must be zero, but yet there is a constant acceleration by the Earth's gravity at that moment. Even though the velocity at that stationary moment is zero, it doesn't imply the acceleration must be zero, so it has a value and in this case is -9.8m/s^{2}

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An alpha particle travels at a velocity of magnitude 550 m/s through a uniform magnetic field of magnitude 0.045T. (An alpha par
creativ13 [48]

Answer:

the force is perpendicular to the speed, it is a type of force that changes the direction of the speed, as in the uniform circular motion te, but does not change its modulus.

Explanation:

The magnetic force is given by the expression

    F = q v x B

The bold are vectors, where v is the velocity and B is the magnetic field, the product is the cross product whose result is a vector perpendicular to the two vectors (v and B)

From the above, the force is perpendicular to the speed, it is a type of force that changes the direction of the speed, as in the uniform circular motion te, but does not change its modulus.

   

Even when the change in direction is real and is caused by a centripetal force

For there to be a change in the velocity modulus there must be a force parallel to the velocity direction, generally a force in electrical

3 0
3 years ago
A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in
otez555 [7]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

 a_{max} = 0.00246 \  m/s^2

b

   k  =722.2 \ N/m

Explanation:

From the question we are told that

     The  amplitude is A  = 1.8 \ cm  =  0.018 \ m

     The period is T  = 17 \ s

    The test weight is  W =  13 \ N

Generally the radial acceleration is mathematically represented as

        a =  w^2 r

at maximum angular acceleration

       r =  A

So  

       a_{max} =  w^2 A

Now w is the angular velocity which is mathematically represented as

      w =  \frac{2 * \pi }{T}

Therefore

       a_{max} =  [\frac{2 *  \pi}{T} ]^2 * A

substituting values

       a_{max} =  [\frac{2 *  3.142}{17} ]^2 * 0.018

       a_{max} = 0.00246 \  m/s^2

Generally this test weight is mathematically represented as

     W =  k *  A

Where k is the spring constant

Therefore

        k  = \frac{W}{A}

substituting values        

      k  = \frac{13}{0.018}

     k  =722.2 \ N/m

7 0
3 years ago
. On a safari, a team of naturalists sets out toward a research station located 9.6 km away in a direction 42° north of east. Af
Elden [556K]

Answer:\theta =49.76^{\circ} North of east

Explanation:

Given

Research station is 9.6 km away in 42^{\circ}North of east

after travelling 3.1 km 25^{\circ} north of east

Position vector of safari after 3.1 km is

r_2=3.1cos25\hat{i}+3.1sin25\hat{j}

Position vector if had traveled correctly is

r_0=9.6cos42\hat{i}+9.6sin42\hat{j}

Now applying triangle law  of vector addition we can get the required vector(r_1)

r_1+r_2=r_0

r_1=(9.6cos42-3.1cos25)\hat{i}+(9.6sin42-3.1sin25)\hat{j}

r_1=4.325\hat{i}+5.112\hat{j}

Direction is given by

tan\theta =\frac{y}{x}=\frac{5.112}{4.325}

\theta =49.76^{\circ}

8 0
3 years ago
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