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2 years ago

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A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc

illates up and down in simple harmonic motion. Calculate the period of motion.

1 answer:

marin [14]2 years ago

4 0

Answer:

Time period of the osculation will be 0.0671 sec

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So $kx=mg$

$k\times 0.04=0.012\times 9.8$

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

$T=2\pi \sqrt{\frac{m}{k}}$

$=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec$

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Answer:

The correct option is;

Force of Friction

Explanation:

As coach Hogue rode his motorcycle round in circle on the wet pavement, the motorcycle and the coach system tends to move in a straight path but due to intervention by the coach they maintain the circular path

The motion equation is

v = ωr and we have the centripetal acceleration given by

α = ω²r and therefore centripetal force is then

m×α = m × ω²r = m × v²/r

The force required to keep the coach and the motorcycle system in their circular path can be obtained by the impressed force of friction acting towards the center of the circular motion.

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3 years ago

What is the net force on this object?

damaskus [11]

Upward and downward forces cancel out. Net force is 8 newtons to the right

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2 years ago

Read 2 more answers

A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v

kifflom [539]

Here we will say that there is no external torque on the system so we will have

$L_i = L_f$

here we know that

$L_i = I_1\omega_1$

where we know that

$I_1 = \frac{2}{5}mr^2$

Also we know that

$I_2 = \frac{1}{12}mL^2$

initial angular speed will be

$\omega_1 = 2\pi(2rev/s) = 4\pi rad/s$

now from above equation

$\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega$

$0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega$

$0.452 = 0.1875 \omega$

now we have

$\omega = 2.41 rad/s$

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2 years ago

2. A test reveals that 150 J of work is required to lift an object 3 m at a

Nuetrik [128]

Answer:

50N

Explanation:

W=Fd

150=F(3)

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2 years ago

The steering wheel of a car has a radius of 0.19 m, and the steering wheel of a truck has a radius of 0.25 m. The same force is

kodGreya [7K]

Answer:

$\frac{T_t}{T_c} = 1.32$

Explanation:

The torque applied on an object can be calculated by the following formula:

$T = Fr$

where,

T = Torque

F = Applied Force

r = radius of the wheel

For car wheel:

$T_c = Fr_c\\$

For truck wheel:

$T_t = Fr_t$

Dividing both:

$\frac{T_t}{T_c} = \frac{Fr_t}{Fr_c}$

for the same force applied on both wheels:

$\frac{T_t}{T_c} = \frac{r_t}{r_c} \\$

where,

rt = radius of the truck steering wheel = 0.25 m

rc = radius of the car steering wheel = 0.19 m

Therefore,

$\frac{T_t}{T_c} = \frac{0.25\ m}{0.19\ m} \\$

$\frac{T_t}{T_c} = 1.32$

8 0

2 years ago