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ASHA 777 [7]
3 years ago
7

A baseball has a mass of 0.145 kg. A professional pitcher throws a baseball 67 mi/h, which is 30.0 m/s. What is the magnitude of

the momentum of the pitched baseball?
Physics
2 answers:
Kamila [148]3 years ago
6 0

The concept of momentum tells us that it is equivalent to the product between the mass and the velocity of the object, that is to say that in general it can be written as

p = mv

Where,

m = mass

v = Velocity

Our values are given as,

m = 0.145kg

v = 67mi/h = 30m/s

Replacing we have that,

p = (0.145)(30)

p = 12.45 kg\cdot m/s

Therefore the magniude of the momentum of the pitched baseball is 12.45 kg\cdot m/s

defon3 years ago
4 0

Answer:

Momentum = 0.145kg × 30m/s = 4.35kgm/s

The momentum of the pitched baseball is 4.35kgm/s

Explanation:

Momentum is the quantity of motion of a moving body, it is the product of mass and velocity of a moving body.

Momentum = mass × velocity = mv

In the case above,

Mass m = 0.145kg

Velocity v = 30m/s

Substituting the values; Momentum will be equal to,

Momentum = 0.145kg × 30m/s = 4.35kgm/s

The momentum of the pitched baseball is 4.35kgm/s

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Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre
castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:  

V

=

d

x

/d

t

=

7.5

×m

i

/

h

r

...........................(

1

)  

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

×=

∫

d

t

+

C

or

−

4.7619  

=

t

+

C

Here C =constant of integration.   

x

=

0  at  t

=

0

, we get:  C

=

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619  =

t

−

4.7619

.

.

.

.

.

.

.

.

.

(

2

)

Here

x  is measured in miles and  t  in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),    

     to get:

      = −

4.7619  

      =  

1

−

4.7619

      = −

3.7619

  or  x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in   m

i

l

/

differentiate  

     equation (1) with respect to time.

     we have to eliminate x from the equation (1) using equation (2).  

     Eliminating x we get:

     v

=

7.5×

     Now differentiating above equation w.r.t time we get:

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=

d

v/

d

t

       =

−

0.675

/

      At  

      t

=

0

      the joggers acceleration is :

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=

−

0.675

m

i

l

/

        =

−

4.34

×

f

t

/  

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        x

=

6  in equation (2).  We get:

        −

4.7619

(

1

−

(

0.04

×

6  )

)^

7

/

10=

t

−

4.7619

         or

         t

=

0.832

h

r

s

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3 years ago
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