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Korvikt [17]
3 years ago
10

What is the acceleration of the object?

Physics
1 answer:
mars1129 [50]3 years ago
4 0
Resultant force is 400-600=-200N.
-200=60a solve for a.
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A 4.00-kg model rocket is launched, expelling 64.0 g of burned fuel from its exhaust at a speed of 675 m/s. What is the velocity
olga_2 [115]

Answer:

The velocity of the rocket is 7.8 m/s

Explanation:

3 0
2 years ago
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
3 years ago
1*2+2+1+5 by how much?
kozerog [31]

Answer:

10

Explanation:

1*2+2+1+5=10

1x2+2+1+5=10

6 0
3 years ago
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:

      a

=

d

v/

d

t

       =

−

0.675

/

      At  

      t

=

0

      the joggers acceleration is :

       a

=

−

0.675

m

i

l

/

        =

−

4.34

×

f

t

/  

(c)  required time for the jogger to run 6 miles is obtained by setting  

        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

6 0
3 years ago
A 75 kV x-ray machine gives an exposure of 200 mR for a 0.2-sec chest x-ray exam.
pashok25 [27]

Explanation:

Given that,

Potential = 75 kV

Exposure = 200 mR

Time = 0.2 sec

We need to calculate the x-ray fluence during this chest x-ray exam

Using formula of fluence

x-ray\ fluence=exposure\times time

Put the value into the formula

x-ray\ fluence=200\times0.2

x-ray\ fluence=40\ mRs

We need to calculate the energy fluence

Using formula of energy fluence

E_{f}=40\times10^{-3}\times75\times10^{3}

E_{f} = 3000\ J

We need to calculate dose -equivalent delivered to the bone, muscle, and fat

Using formula of dose

D=\dfrac{E}{t}

Where, D = dose

E = energy

t = time

Put the value into the formula

D=\dfrac{3000}{0.2}

D=15.0\ kJ

Hence, This is the required solution.

7 0
3 years ago
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