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storchak [24]
2 years ago
11

A block with mass m = 4.3 kg is attached to two springs with spring constants kleft = 35 N/m and kright = 48 N/m. The block is p

ulled a distance x = 0.23 m to the left of its equilibrium position and released from rest.What is the magnitude of the net force on the block (the moment it is released)?
Physics
1 answer:
Mrrafil [7]2 years ago
4 0

Answer:

Explanation:

A restoring force is created when a spring is either stretched or compressed.

Restoring force created by first spring = spring constant x contraction

= 35 x .23

= 8.05 N .

It will act towards the right .

Restoring force created by second spring = spring constant x extension

= 48 x .23

= 11.04 N .

It will also act towards the right .

Total force created due to contraction in the first spring and extension in the right spring = 8.05  +11.04 N

= 19.09 N .

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The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th
worty [1.4K]

Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg

where m_a is the mass of the arrow, m_b is the mass of the block, \Delta H of the change in height of the block after the collision, and v is the velocity of the arrow before it hit the block.

Solving for the velocity v, we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} } $

and we put in the numerical values

m_a = 0.045kg,

m_b = 1.40kg,

\Delta H = 0.4m,

g= 9.8m/s^2

and simplify to get:

\boxed{ v= 15.9m/s}

The arrow was moving at 15.9 m/s

6 0
3 years ago
How much work is done against gravity when lowering a 16 kg box 0.50 m? (g = 9.8 m/s2)
leonid [27]

Answer:

The work done against gravity is 78.4 J

Explanation:

The work is calculated by multiplying the force by the distance that the

object moves

W = F × d, where W is the work , F is the force and d is the distance

The SI unit of work is the joule (J)

We need to find the work done against gravity when lowering a

16 kg box 0.50 m

→ F = mg

→ m = 16 kg, and g = 9.8 m/s²

Substitute these value in the rule

→ F = 16 × 9.8 = 156.8 N

→ W = F × d

→ F = 156.8 N and d = 0.50

Substitute these values in the rule

→ W = 78.4 J

<em>The work done against gravity is 78.4 J</em>

6 0
3 years ago
Which of the following statements is true regarding the constructive interference diagram shown below? Select all that apply.
yulyashka [42]

Answer:

I believe its B and C

Explanation:

<u><em>If I'm wrong please tell me so I can correct my answer.</em></u>

<u><em /></u>

6 0
2 years ago
Read 2 more answers
The average lifetime of μ-mesons with a speed of 0.95c is measured to be 6 x 10^6 s. Find the average lifetime of μ-mesons in a
Mnenie [13.5K]

Answer:

19.2*10^6 s

Explanation:

The equation for time dilation is:

t = \frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}

Then, if it is observed to have a life of 6*10^6 s, and it travels at 0.95 c:

t = \frac{6*10^6}{\sqrt{1-\frac{(0.95c)^2}{c^2}}} = 19.2*10^6 s

It has a lifetime of 19.2*10^6 s when observed from a frame of reference in which the particle is at rest.

7 0
3 years ago
Barney walks at a velocity of 1.7 meters/second on an inclined plane, which has an angle of 18.5° with the ground. What is the h
Viktor [21]
The velocities and the speed build a triangle, where the 1.7 m/s are the hypotenuse and the x-velocity and y-velocity are the other sides. 

<span>So the x-velocity is: speed*cos(angle) </span>

<span>now plug in </span>
<span>x=1.7 m/s * cos(18.5)=1.597 m/s </span>


3 0
2 years ago
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