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kari74 [83]
2 years ago
8

B. Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

Physics
1 answer:
goblinko [34]2 years ago
7 0

We have that the instantaneous velocity of the shuttlecock when it hits the ground is

V_{int}=\sqrt{U^2+19.6H}

From the question we are told

Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

shuttlecock when it hits the ground? Show your work below.

Generally the equation for acceleration  is mathematically given as

a=v \frac{dv}{dx}\\\\\Therefore\\\\\2ah=v^2-u^2

Where

acceleration is still -9.81 m/s2,

Hence,

V^2-U^2=2(-9.81)*-H

Therefore

V_{int}=\sqrt{U^2+19.6H}

For more information on this visit

brainly.com/question/12319416?referrer=searchResults

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Solnce55 [7]

La respuesta correcta para esta pregunta abierta es la siguiente.

No incluyes opciones, incisos, o alguna referencia específica para responder esta pregunta.

De ahí que responderemos en términos generales basándonos en nuestro conocimiento previo.

El concepto cosmológico abarca todo lo relativo a la cosmología y, al unir este término con la metafísica sería posible una correcta asociación la cual sería benéfica para ambos campos de conocimiento porque de esta manera se estarían complementando y enriqueciendo el uno al otro.

Cuando la ciencia -en este caso, la cosmología- acepte los conceptos metafísico que no tienen un sustento científico pero que sí contemplan alternativas en el mundo espiritual que son aceptadas por una innumerable cantidad de personas, entonces los expertos podrán plantear nuevas y diferentes posibilidades de entender los fenómenos que al día de hoy, carecen de una convincente explicación.

Tanto la cosmología como la metafísica podrían mostrarse más receptivos a las aportaciones de una y la otra, con objeto de poder aumentar las posibilidades de comprender un fenómeno. No necesariamente debe haber una teoría científica para poder avanzar en el conocimiento de algún tema. De ahí que las aportaciones metafísicas sean válidas como una alternativa de investigación.

5 0
2 years ago
A spaceship is travelling at 20,000.0 m/s. After 5.0 seconds, the rocket thrusters are turned on. At the 55.0 second mark, the s
tankabanditka [31]

Answer:

80 m/s^2

Explanation:

The acceleration of an object is given by:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval it takes for the velocity to change from u to v

For the rocket in this problem,

u = 20,000 m/s

v = 24,000 m/s

t = 55.0 - 5.0 = 50.0 s

Substituting,

a=\frac{24000-20000}{50}=80 m/s^2

7 0
2 years ago
The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

3 0
3 years ago
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Use the circuit diagram to decide if the lightbulb will light. Justify your answer.
podryga [215]

Answer:

The lightbulb will NOT light.

Explanation:

You put me in a difficult position.  I can't help it, but the "sample answer" is by far the best way to explain this, briefly and correctly.  There's no other choice but to copy it.

This is a short circuit. The branch without the bulb has almost no resistance, so all the current will flow through that branch instead of flowing through the bulb.

<em>If</em> the lower switch were <u>opened</u>, THEN we would have a series circuit.  Current would no longer have any other choice but to flow through the bulb, and the bulb would light.

5 0
3 years ago
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A lamp draws a current of 0.50 A when it is connected to a 120 V source? What is the resistance of the lamp?
emmainna [20.7K]
Given,
Current (I) = 0.50A
Voltage (V) = 120 volts
Resistance (R) =?
We know that:-
Voltage (V) = Current (I) x Resistance (R)
→Resistance (R) = Voltage (V) / Current (I)
= 120/0.50
= 24Ω
∴ Resistance (R) = 24Ω
8 0
3 years ago
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