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Alisiya [41]
4 years ago
13

1. How is acceleration different from velocity?

Physics
1 answer:
Drupady [299]4 years ago
8 0

Answer: The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Whereas, acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.

-Here is a formula for acceleration.

-And due note there are multiple types of velocity, namely starting and final velocity there are multiple formulas. I urge you to learn them!

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You stand on top a building 44 m tall with a water balloon. You drop the water balloon from rest. How fast is the balloon moving
Alecsey [184]
<h2>The balloon is moving when it is halfway down the building at 20.78 m/s.</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 0 m/s  

Acceleration, a = 9.81 m/s²  

Displacement, s = 0.5 x 44 = 22 m

Substituting  

v² = u² + 2as

v² = 0² + 2 x 9.81 x 22

v² = 431.64

v = 20.78 m/s

Velocity at 22 m = 20.78 m/s

The balloon is moving when it is halfway down the building at 20.78 m/s.

7 0
3 years ago
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x= 5.0 cos
lakkis [162]

Answer:

a)   x = 4.33 m ,   b)  w = 2 rad / s ,  f = 0.318 Hz ,  c) a = - 17.31 cm / s²,  

d) T =  3.15 s,  e)  A = 5.0 cm

Explanation:

In this exercise on simple harmonic motion we are given the expression for motion

          x = 5 cos (2t + π / 6)

they ask us for t = 0

a) the position of the particle

      x = 5 cos (π / 6)

      x = 4.33 m

remember angles are in radians

 

b) The general form of the equation is

          x = A cos (w t + Ф)

when comparing the two equations

         w = 2 rad / s

angular velocity and frequency are related

          w = 2π f

           f = w / 2π

           f = 2 / 2pi

           f = 0.318 Hz

c) the acceleration is defined by

      a == d²x / dt²

      a = - A w² cos (wt + Ф)

for t = 0 ,  we substitute

      a = - 5,0 2² cos (π / 6)

      a = - 17.31 cm / s²

d) El period is

          T = 1/f

         T= 1/0.318

         T =  3.15 s

e) the amplitude

        A = 5.0 cm

3 0
3 years ago
In just 0.30 s , you compress a spring (spring constant 5000 n/m ), which is initially at its equilibrium length, by 4.0 cm. par
Ierofanga [76]
The average power output is the ratio between the work done to compress the spring, W, and the time taken, t:
P= \frac{W}{t} (1)

The work done is equal to the elastic energy stored by the compressed spring:
W=U= \frac{1}{2}kx^2
where k=5000 N/m is the spring constant and x=4.0 cm=0.04 m is the compression of the spring. If we substitute the numbers, we find:
W= \frac{1}{2}(5000 N/m)(0.04 m)^2=4 J

And now we can use eq.(1) to calculate the average power output:
P= \frac{W}{t}= \frac{4 J}{0.30 s}  =13.3 W
7 0
3 years ago
A batter hits a baseball so that it leaves the bat with an initial speed of 60m/s at an initial angle of 42 with the horizontal
Elza [17]

Answer: 24.06°

Explanation:

So,we can say after t if it reaches height h then,

h = (37 sin 53)t - 1/2 * 9.8t^2 (as,vertical component of velocity is 37 sin 53)

Given t = 2s

So, h = 39.5m

And horizontal displacement will be

r = 37 cos 53 *2 = 44.52m

So,after 2s the baseball will be lying 39.57m

above its point of projection and 44.52m ahead of its point of projection.

Now.let the vertical component of velocity will become Vy after time 2s

So, Uy = 37 sin 53- 9.8* 2

or, U = 9.95m/s

And.horizontal component of velocity remains

constant i.e Vx = 37 cos 53 = 22.27m/s

So.magnitude of velocity after 2s is

Square root of (Vx^2 + Vy^2)= 24.4m/s

Making an angle of tan 22.27/9.95 = 24.06°

6 0
3 years ago
The objective lens and the eyepiece of a telescope are spaced 85 cm apart. If the eyepiece is123 D what is the total magnificati
dedylja [7]

Answer:

The magnification would be "103.55". A further explanation is given below.

Explanation:

The given values are:

Distance between lens and eyepiece,

L = 85 cm

Eyepiece is,

= 123 D

Now,

The refractive power of eye piece will be:

⇒ \frac{1}{f_e}=123D

   f_e=\frac{1}{123D}

   f_e=0.813 \ cm

The length of the telescope will be:

⇒ L=f_0+f_e

⇒ f_0=L-f_e

On substituting the values, we get

⇒     =85-0.813

⇒     =84.187 \ cm

Now,

The magnification of the telescope will be:

⇒ M=\frac{f_0}{f_e}

⇒      =\frac{84.187}{0.813}

⇒      =103.55

5 0
3 years ago
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