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Evgen [1.6K]
3 years ago
7

In a test strength at a carnival, a hammer drives a metal weight up a rod to ring a bell at the top. If the bell is 24 ft above

the starting position of the weight, what is the minimum initial vertical velocity needed to ring the bell?

Physics
1 answer:
vladimir2022 [97]3 years ago
6 0

Answer:

39 ft/s

Explanation:

Given:

Δy = 24 ft

v = 0 ft/s

a = -32 ft/s²

Find: v₀

v² = v₀² + 2aΔy

(0 ft/s)² = v₀² + 2 (-32 ft/s²) (24 ft)

v₀ ≈ 39 ft/s

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A 57 g ice cube can slide without friction up and down a 33 ∘ slope. The ice cube is pressed against a spring at the bottom of t
beks73 [17]
The springs stored energy is transferred to the cube as kinetic energy and then by the slop the KE is converted to height energy. 

<span>0.5 . k . x^2 = 0.5 . m . v^2 = m . g . ∆h </span>

<span>0.5 . 50 . (0.1^2) = 0.05 . 9.8 . ∆h </span>

<span>∆h = 0.51 m = 51 cm </span>

<span>This is the height gained </span>
<span>Distance along the slope = ∆h / sin 60 = 0.589 = 59 cm </span>

<span>In the second case, the stored spring energy is converted into height energy AND frictional heat energy. </span>

<span>The height energy is m . g . d sin 60 where d is the distance the cube moves along the slope. </span>

<span>The Frictional energy converted is F . d </span>

<span>F ( the frictional force ) = µ . N </span>

<span>N ( the reaction to the component of the gravity force perpendicular to the surface of the slope ) = m . g . cos60 </span>

<span>Total energy converted </span>

<span>0.5 . k . x^2 = (m . g . dsin60) + (µ . m . g . cos60 . d ) </span>

<span>Solve for d </span>

<span>d = 0.528 = 53 cm</span>
5 0
3 years ago
What factor causes atmospheric pressure
bagirrra123 [75]
Atmospheric pressure is caused by the weight of the atmosphere pushing down on itself and on the surface below it.
Pressure is defined as the force acting on an object divided by the area upon witch the force is acting.
7 0
3 years ago
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Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and
lawyer [7]

1 kilometre=1000 metre

      1 hour = 3600 second

       1\ km/hr=\frac{1000}{3600} m/s

       1\ km/hr=\frac{5}{18} m/s

The initial velocity of car A is 35.0 km/hr i.e

                                         35.0\ km/hr=35*\frac{5}{18} m/s

                                                                   = 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as  25\ km/hr^2

                                            =\ 25*\frac{1000}{3600*3600} m/s^2

                                            =0.00192901234 m/s^2

The time taken by car A = 15 min.

From equation of kinematics we know that-

                                 v= u+at      [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A,  v = 9.72 m/s +[0.00192901234×15×60]m/s

                                   =11.456111106 m/s

The acceleration of B is given as    15\ km/hr^2

                                    =0.00115740740740 m/s^2

The time taken by car B =20 min

The final velocity of B is -

                             v= u+at

                               = u-at    [Here a is negative due to deceleration]

                               =12.5 m/s +[0.0011574074074×20×60]

                               =13.8888888.....

                               =13.9

The acceleration of C is given as    40\ km/hr^2          

                                                            =\ 0.003086419753 m/s^2

The time taken by car C =30 min

The final velocity of C is-

                                v = u+at

                                   =8.89 m/s+[0.003086419753×30×60] m/s

                                   =14.4455555555..m/s

                                   =14.45 m/s

The car C is decelerating.The deceleration is given as-  60\ km/hr^2

                                                                      =0.0046296296296m/s^2

The time taken by car D= 45 min.

The final velocity of the car D is-

                     v =u+at

                        =30.56 -[0.00462962962962×45×60]m/s

                        =18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

                 


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3 years ago
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Answer:

b

Explanation:

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3 years ago
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Hans observed properties of four different waves and recorded observations about each one in his chart.
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The answer to your question is (Wave W is being diffracted, Wave X is being reflected, and waves Y and Z are being refracted.)


HOPE THIS HELPS YOU CAUSE IM DOING TEST AND I HOPE IM RIGHT AS WELL

PLZ GIVE ME THE BRAINLIEST ANSWER HAVE A GOOD DAY




8 0
3 years ago
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