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3 years ago

12

Question 1

1 answer:

galina1969 [7]3 years ago

7 0

1/f= 1/di+1/do
1/f=1/90m+1/40m
1/f ≈27.7m

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PtichkaEL [24]

Answer:

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4 0

3 years ago

The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it

MrRissso [65]

Answer:

Explanation:

From the question we are told that

The moment of inertia is $I = 2.50 \ kg \cdot m^2$

The final angular speed is $w_f = 400 rev/min = \frac{400 * 2\pi}{60} = 41.89 \ rad/s$

The time taken is $t = 8.0 s$

The initial angular speed is $w_i = 0\ rad/s$

Generally the average angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_i }{t}$

=> $\alpha = \frac{41.89}{8}$

=> $\alpha = 5.24 \ rad/s^2$

Generally the torque is mathematically represented as

$\tau = I * \alpha$

=> $\tau = 5.24 * 2.50$

=> $\tau = 13.09 \ N \cdot m$

5 0

3 years ago

Can a nickel be attracted to a magnet?

zhuklara [117]

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8 0

3 years ago

Why are all bodies in nature electrically neutral? short answer please

DENIUS [597]

Answer: because they have equal numbers of protons

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3 0

3 years ago

A firecracker breaks up into two pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/

larisa [96]

Answer:

Total momentum, p = 21.24 kg-m/s

Explanation:

Given that,

Mass of first piece, $m_1=200\ g= 0.2\ kg$

Mass of the second piece, $m_2=300\ g= 0.3\ kg$

Speed of the first piece, $v_1=82\ m/s$ (along x axis)

Speed of the second piece, $v_2=45\ m/s$ (along y axis)

To find,

The total momentum of the two pieces.

Solve,

The total momentum of two pieces is equal to the sum of momentum along x axis and along y axis.

$p_x=m_1v_1$

$p_x=0.2\ kg\times 82\ m/s$

$p_x=16.4\ kg-m/s$

$p_y=m_2v_2$

$p_y=0.3\ kg\times 45\ m/s$

$p_y=13.5\ kg-m/s$

The net momentum is given by :

$p=\sqrt{p_x^2+p_y^2}$

$p=\sqrt{16.4^2+13.5^2}$

p = 21.24 kg-m/s

Therefore, the total momentum of the two pieces is 21.24 kg-m/s.

4 0

3 years ago