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3 years ago

Which part provides the resistance force in the

Physics

2 answers:

Aleonysh [2.5K]3 years ago

3 0

Answer:

D- clay block

Explanation:

tensa zangetsu [6.8K]3 years ago

3 0

Answer:

The correct answer is D.

Explanation:

The clay block provides the force of resistance in the photo, since the weight it exerts on the right end of the ruler on which it is supported makes the system stay in balance. The weight of the spring balance (object A) is transferred to its point of support on the ruler, while the wood and clay cube only exerts one point of support, but does not exert any force on the ruler. The weight exerted by the clay block downward counteracts the weight of the spring balance.

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4 0

2 years ago

A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to ro

NARA [144]

Answer:

the required minimum magnitude of the force F is 21 N

Explanation:

Given the data in the question,

m = 5 kg

width = 60 cm

height = 80 cm

Let force is F represent in the image below,

so when the block about to rotate normal shifted to edge of cube

mg(w/2) = Fh

F = mg(w/2) / h

we know that g = 9.8 m/s²

we substitute

F = (5 × 9.8 ( 60/2)) / 70

F = (5 × 9.8 × 30 ) / 70

F = 1470 / 70

F = 21 N

Therefore, the required minimum magnitude of the force F is 21 N

5 0

2 years ago

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$