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expeople1 [14]
3 years ago
10

The distance a spring stretches varies directly as the amount of weight that is hanging on it. A weight of 2.5 pounds stretches

a spring 18 inches. Find the stretch of the spring when a weight of 6.4 pounds is hanging on it.
Physics
2 answers:
julsineya [31]3 years ago
5 0
We use the equation y = kx
y = 18, x = 2.5

18 = k2.5
k = 7.2

x = 6.4
y = 7.2*6.4
y = 46.08

hope this help
V125BC [204]3 years ago
3 0

Answer:

The stretch of the spring is 46.08 inches.

Explanation:

Given that,

Weight = 2.5 pounds

Stretches a spring = 18 inches

new weight = 6.4

We need to calculate the stretch of the spring

let stretch of the spring is x .

Using formula of restoring force

F=-kx

For first weight,

2.5=k\times18....(I)

For another weight,

6.4=k\times x....(II)

Ratio of equation (I) and (II)

\dfrac{2.5}{6.4}=\dfrac{18}{x}

x=\dfrac{18\times6.4}{2.5}

x = 46.08\ inches

Hence, The stretch of the spring is 46.08 inches.

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You are launching a 2 kg potato out of a potato cannon. The cannon is 2.0 m long and is aimed 70 degrees above the horizontal. I
DochEvi [55]

Answer:

Explanation:

The net force on the potatoes is given by:

F= 52 - mgSintheta

F= 52- (2×9.8× Sin70°)

F = 52 -18.4

F= 33.58N

Using Newton's 2nd law

F = ma

a=F/m = 33.58/ 2 = 16.79m/s^2

Using the equation of motion:

V^2= u^2 + 2as

V^2 = 0 + 2× 16.79 x2

V^2 = 67.16

V=sqrt(68.16)

V= 8.195m/s This is the exit velocity of the potatoes

Kinetic energy, K.E = 1/2mv^2

KE= 1/2 × 2 × 8.195^2

KE = 67.16J

8 0
3 years ago
Which factors could be potential sources of error in the experiment? Check all that apply.
mote1985 [20]

Answer:

1. energy lost in the lever due to friction

3. visual estimation of height of the beanbag

5. position of the fulcrum for the lever affecting transfer of energy

Explanation:

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8 0
3 years ago
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3 years ago
Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at
beks73 [17]

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

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8 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
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