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2 years ago

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9. A 12 v battery is connected to four 5 ohm light bulbs. What is the equivalent

1 answer:

vlabodo [156]2 years ago

7 0

20 ohms 5 ohms
12volts

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___________ force can be thought of as either a push or a pull, as long as it makes the object move in a circular path.

fomenos

I believe it would be the Centripetal force.

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2 years ago

Read 2 more answers

The gravitational force between two objects has a magnitude of F. If both masses were doubled and the distance between them doub

Fofino [41]

Answer:

F' = F

Explanation:

The gravitational force of attraction between two objects can be given by Newton's Gravitational Law as follows:

$F = \frac{Gm_1m_2}{r^2}$

where,

F = Force of attraction

G = Universal gravitational costant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects

Now, if the masses and the distance between them is doubled:

$F' = \frac{G(2m_1)(2m_2)}{(2r)^2}\\\\F' = \frac{Gm_1m_2}{r^2}$

<u>F' = F</u>

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2 years ago

Based on the table above, _______ planets are made of the same things as _______ moons.

wolverine [178]

Answer:

b

Explanation:

3 0

2 years ago

Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the c

Amanda [17]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is $v_A= 4m/s$

Explanation:

From the question we are told that

The length of the string is $L = 1m$

The initial speed of block A is $u_A$

The final speed of block A is $v_A = \frac{1}{2}u_A$

The initial speed of block B is $u_B = 0$

The mass of block A is $m_A = 7kg$ gh

The mass of block B is $m_B = 2 kg$

According to the principle of conservation of momentum

$m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}$

Since block B at initial is at rest

$m_A u_A = m_Bv_B + m_A \frac{u_A}{2}$

$m_A u_A - m_A \frac{u_A}{2} = m_Bv_B$

$m_A \frac{u_A}{2} = m_Bv_B$

making $v_B$ the subject of the formula

$v_B =m_A \frac{u_A}{2 m_B}$

Substituting values

$v_B =\frac{7 u_A}{4}$

This $v__B$ is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of $v__B'$ at the top of the vertical circle

The angular centripetal acceleration would be mathematically represented

$a= \frac{v^2_{B}'}{L}$

Note that this acceleration would be toward the center of the circle

Now the forces acting at the top of the circle can be represented mathematically as

$T + mg = m \frac{v^2_{B}'}{L}$

Where T is the tension on the string

According to the law of energy conservation

The energy at bottom of the vertical circle = The energy at the top of

the vertical circle

This can be mathematically represented as

$\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L$

From above

$(T + mg) L = m v^2_{B}'$

Substitute this into above equation

$\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = T + 5mgL$

The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero

This is mathematically represented as

$\frac{49 mv_A^2}{16} = 5mgL$

making $v_A$ the subject

$v_A = \sqrt{\frac{80mgL}{49m} }$

substituting values

$v_A = \sqrt{\frac{80* 9.8 *1}{49} }$

$v_A= 4m/s$

6 0

2 years ago

Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid ra

BabaBlast [244]

Explanation:

It is given that,

Length of the helicopter, l = 3.1 m

The helicopter rotates, the length of helicopter will become the radius of circular path, r = 3.1 m

Angular speed of the helicopter, $\omega=280\ rev/min=29.32\ rad/s$

(a) The centripetal acceleration in terms of angular velocity is given by :

$a_c=r\times \omega^2$

$a_c=3.1\times (29.32)^2$

$a_c=2664.95\ m/s^2$

(b) Let v is the linear speed of the tip. The relation between the linear and angular speed is given by :

$v=r\times \omega$

$v=3.1\times 29.32$

v = 90.89 m/s

$\dfrac{v_{tip}}{v_{sound}}=\dfrac{90.89}{340}=0.267$

Hence, this is the required solution.

4 0

2 years ago