All categories

3 years ago

Need help with the question #14

Physics

1 answer:

Ivanshal [37]3 years ago

6 0

BEFORE the shot, the cannon is just sitting there with the ball inside.
Their combined momentum is zero.

Momentum is conserved, so their combined momentum has to be
zero AFTER the shot too.

Whatever momentum the ball has in one direction, the cannon
must have the same momentum in the other direction.

Momentum = (mass) x (speed)

Momentum of the ball = (6 kg) x (200 m/s) = 1,200 kg-m/s

Momentum of the cannon =

(2,000 kg) x (speed) = 1,200 kg-m/s

Divide each side by (2,000 kg)

speed = (1,200 kg-m/s) / (2,000 kg)

= 0.6 m/s

That's where the recoil of any gun comes from. The same
momentum as the bullet has, but in the opposite direction.

You might be interested in

What family of elements has 7 valence electrons?

ivann1987 [24]

Group 17 (VII) (halogens)

4 0

3 years ago

Read 2 more answers

Give one real-life example of the Doppler Effect and explain what causes this phenomenon. Please write at least 3 complete sente

Elanso [62]

Dipper effect of an oncoming train get louder as it approaches and sound diminishes as it goes away sound traveling

7 0

3 years ago

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.85 kg and length L = 5.76 m to a uniform sphere with

Ket [755]

Answer:

Part a)

$I = 1879.7 kg m^2$

Part b)

$\alpha = 0.70 rad/s^2$

Part c)

$I = 153.8 kg m^2$

Part 4)

angular acceleration will be ZERO

Part 5)

$I = 345.6 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about left end of the rod is given as

$I = \frac{m_r L^2}{3} + (\frac{2}{5} m_s R^2 + m_s(R + L)^2)$

So we have

$I = \frac{m_r(4R)^2}{3} + (\frac{2}{5}(5m_r) R^2 + (5m_r)(R + 4R)^2)$

$I = \frac{16}{3}m_r R^2 + (2m_r R^2 + 125 m_rR^2)$

$I = (\frac{16}{3} + 127)m_r R^2$

$I = (\frac{16}{3} + 127)(6.85)(1.44)^2$

$I = 1879.7 kg m^2$

Part b)

If force is applied to the mid point of the rod

so the torque on the rod is given as

$\tau = F\frac{L}{2}$

$\tau = 460(2R)$

$\tau = 460 \times 2 \times 1.44$

$\tau = 1324.8 Nm$

now angular acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{1324.8}{1879.7}$

$\alpha = 0.70 rad/s^2$

Part c)

position of center of mass of rod and sphere is given from the center of the sphere as

$x = \frac{m_r}{m_r + m_s}(\frac{L}{2} + R)$

$x = \frac{m_r}{6 m_r}(3R) = \frac{R}{2}$

so moment of inertia about this position is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + \frac{R}{2})^2 + (\frac{2}{5} m_s R^2 + m_s(\frac{R}{2})^2)$

so we have

$I = \frac{m_r (16R^2)}{12} + m_r(\frac{5R}{2})^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(\frac{R^2}{4})$

$I = m_r R^2(\frac{16}{12} + \frac{25}{4} + 2 + \frac{5}{4})$

$I = 6.85(1.44)^2\times 10.83$

$I = 153.8 kg m^2$

Part 4)

If force is applied parallel to the length of rod

then we have

$\tau = \vec r \times \vec F$

$\tau = 0$

so angular acceleration will be ZERO

Part 5)

moment of inertia about right edge of the sphere is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + 2R)^2 + (\frac{2}{5} m_s R^2 + m_s(R)^2)$

so we have

$I = \frac{m_r (16R^2)}{12} + m_r(4R)^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(R^2)$

$I = m_r R^2(\frac{16}{12} + 16 + 2 + 5)$

$I = 6.85(1.44)^2\times 24.33$

$I = 345.6 kg m^2$

6 0

3 years ago

Why is energy transferable?

disa [49]

Kinetic Energy. Energy is transferred from one object to another when a reaction takes place. Energy comes in many forms and can be transferred from one object to another as heat, light, or motion, to name a few. ... This energy would be in the form of motion, with the person lifting the blue ball to a higher level.

3 0

3 years ago

If the mass is 22 g and the volume is 11 ml, what is the density of the object

andriy [413]

2 g/mL there’s your answer

8 0

3 years ago

Read 2 more answers