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2 years ago

Can you answer these questions please

Physics

2 answers:

wolverine [178]2 years ago

6 0

Can you type the words, and questions? I will answer once you do. <3

Lelechka [254]2 years ago

6 0

just explain correctdomondo

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A diagram of a closed circuit with power source on the left labeled 12 V, a resistor on the top labeled 10 Ohms, a resistor on t

zalisa [80]

Answer:

7.5 A

6.0 A

5.0 A

Explanation:

6 0

3 years ago

Read 2 more answers

While finding the spring constant, if X1 = 12 cm, X2 = 15 cm, and hanging mass = 22 grams, the value of spring constant K would

Pavel [41]

Answer:

If x₁=12 cm then k=1.7985 N/m

If x₂=15 cm then k=1.4388 N/m

Explanation:

Hanging mass= 22 g=0.022 kg

Acceleration due to gravity g=9.81 m/s²

If x₁=displacement= 12 cm=0.12 m

k= spring constant

$F=ma\\\Rightarrow F=0.022\times 9.81\\\Rightarrow F=0.21582\ N$

$\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.012\\\Rightarrow k=1.7985\ N/m\\$

∴k = 1.7985 N/m

If x₂=15 cm=0.15 m

Force of the hanging mass is same however the spring constant will change

$\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.015\\\Rightarrow k=1.4388\ N/m\\$

∴k = 1.4388 N/m

As the mass is not changing the spring constant has to change. That means that here there are two spring one with k=1.7985 N/m and the other with k= 1.4388 N/m

4 0

2 years ago

An incompressible fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pip

OverLord2011 [107]

Answer:

The value is $v_2 = 5.53 \ m /s$

Explanation:

From the question we are told

The pipe diameter at location 1 is $d = 8.8 \ cm = \frac{8.8 }{10} = 0.88 \ m$

The velocity at location 1 is $v_1 = 2.4 \ m /s$

The diameter at location 2 is $d_2 = 5.80 \ cm = 0.58 \ m$

Generally the area at location 1 is

$A_1 = \pi * \frac{d^2}{ 2}$

=> $A_1 = \pi * \frac{0.88^2}{ 2}$

=> $A_1 = 3.142 * \frac{0.88^2}{ 2}$

=> $A_1 = 1.2166 \ m^2$

Generally the area at location 1 is

$A_2 = \pi * \frac{d_1^2}{ 2}$

=> $A_2 = \pi * \frac{0.58^2}{ 2}$

=> $A_2 = 0.528 \ m^2$

Generally from continuity equation we have that

$A_1 * v_1 = A_2 * v_2$

=> $1.2166 * 2.4 = 0.528 * v_2$

=> $v_2 = 5.53 \ m /s$

3 0

2 years ago

A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s

pashok25 [27]

Answer:

109656.25 Nm

Explanation:

$\omega_f$ = Final angular velocity = 1.5 rad/s

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm$

The torque specifications must be 109656.25 Nm

5 0

2 years ago

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Lelechka [254]

Answer:

a. 16 s b. -1.866 kJ

Explanation:

a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.

We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².

Since the turntable stops at ω = 0, the time it takes to stop is gotten from

ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.

So it takes the turntable 16 s to stop.

b. The workdone by the turntable to stop W equals its rotational kinetic energy change.

So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ

3 0

2 years ago