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3 years ago

Velocity has both speed and direction , so it is a

Physics

1 answer:

Savatey [412]3 years ago

3 0

Answer:

Velocity is a measure of both speed and direction of motion. Velocity is a vector, which is a measurement that includes both size and direction. Velocity can be represented by an arrow, with the length of the arrow representing speed and the way the arrow points representing direction.

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Explanation:

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How to find moment of inertia of hemisphere

gtnhenbr [62]

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3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

3 years ago

How should you behave while performing laboratory experiments?

Brrunno [24]

Explanation:

Always behave responsibly in the laboratory. Do not run around or play practical jokes. Always check the safety data of any chemicals you are going to use. Never smell, taste or touch chemicals unless instructed to do so.

4 0

3 years ago

Read 2 more answers

(URGENT) A ball rolls off a ledge. Its velocity is 7.70m/s in a horizontal direction. It falls on the floor 1.60m below the ledg

elena-s [515]

Answer:

the horizontal distance is 4.355 meters

Explanation:

The computation of the horizontal distance while travelling in the air is shown below:

Data provided in the question is as follows

Velocity = u = 7.70 m/s

H = 1.60 m

R = horizontal direction

Based on the above information

As we know that

R = u × time

where,

Time = $\sqrt{\frac{2H}{g} }$

So,

= $7.70\times \sqrt{\frac{2\times 1.60}{10} }$

= 4.355 meters

hence, the horizontal distance is 4.355 meters

6 0

3 years ago

[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r

blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity $V_{f}$ for an object traveling distance h taking the initial vertical component of velocity as $V_{i}$ the kinematics equation is written as