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Anna11 [10]
2 years ago
10

A 1.0 kg block is attached to an unstretched

Physics
1 answer:
KonstantinChe [14]2 years ago
4 0

Answer:

Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = 1 Nm.

Explanation:

Here, spring constant, k  = 50 N/m.

given block comes down eventually 0.2 m below.

here, g = 10 m/s.

let block be at a height h above the ground in figure 1.

⇒In figure 2, potential energy of the block-spring-Earth

system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.

⇒ Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)

              =  (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.

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You've always wondered about the acceleration of the elevators in the 101 story-tall Empire State Building. One day, while visit
love history [14]

To develop this problem we will proceed to convert all units previously given to the international system for which we have to:

140 lb = 63.5 kg \rightarrow 63.5kg (9.8m/s) =622.3 N

120 lb = 54.4 kg \rightarrow 54.4kg (9.8m/s)= 533 N

170 lb = 77.1 kg \rightarrow 77.1 kg (9.8m/s) =756 N

PART A ) From the given values the minimum acceleration will be given for 120Lb and maximum acceleration when 170Lb is reached therefore:

F = 756 - 622.3

F = 133.7N

Through the Newtonian relationship of the Force we have to:

F= ma

a = \frac{F}{m}

a = \frac{133.7}{63.5}

a = 2.1m/s^2

PART B) For the maximum magnitude of the acceleration downward we have that:

F = 622.3 - 533

F = 89.3N

Through the Newtonian relationship of the Force we have to:

F= ma

a = \frac{F}{m}

a = \frac{89.3}{63.5}

a = 2.1m/s^2

a = 1.04 m/s^2

7 0
2 years ago
PLEEEEEAAASSSEEE HELP ME
Xelga [282]

Answer:

1) A.  0.44 m/s East + 0.33 m/s North

2) A.  0 m/s²

3) A.  a scalar calculated as distance divided by time.

4) B.  31 km per hour

Explanation:

1) Velocity is DISPLACEMENT over time.

at 1 m/s, total time of walking is 9000 seconds

displacement is 3000 m north and 5000 - 1000 = 4000 m east

4000 m/ 9000 s = 0.44 m/s E

3000 m/ 9000s = 0.33 m/s N

2) constant speed means no acceleration

3) A. a scalar calculated as distance divided by time.

4) displacement 50 km N and 80 km W

v = √(50² + 80²) / (1 + 2) = 31.446603... km/hr

3 0
1 year ago
Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68
den301095 [7]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, y_0=1500\ m

Initial horizontal velocity is, u_x=688\ m/s

Initial vertical velocity is, u_y=0(\textrm{Horizontal velocity only initially)}

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is S=y-y_0=0-1500=-1500\ m

Acceleration in the vertical direction is due to gravity, g=-9.8\ m/s^2

Therefore, the displacement of the bomb is given as:

S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s

So, the bomb will remain in air for 17.5 s before hitting the ground.

6 0
2 years ago
A 145-g baseball is thrown so that it acquires a speed of 25 m/s. What was the net work done on the ball to make it reach this s
inysia [295]

When the ball has left your hand and is flying on its own, its kinetic energy is

KE = (1/2) (mass) (speed²)

KE = (1/2) (0.145 kg) (25 m/s)²

KE = (0.0725 kg) (625 m²/s²)

<em>KE = 45.3 Joules</em>

If the baseball doesn't have rocket engines on it, or a hamster inside running on a treadmill that turns a propeller on the outside, then there's only one other place where that kinetic energy could come from:  It MUST have come from the hand that threw the ball.  The hand would have needed to do  <em>45.3 J</em>  of work on the ball before releasing it.

6 0
2 years ago
A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average
frez [133]
 (u) = 20 m/s 
(v) = 0 m/s 
<span> (t) = 4 s 
</span>
<span>0 = 20 + a(4) 

</span><span>4 x a = -20 
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>
8 0
2 years ago
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