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Effectus [21]
1 year ago
12

A 1000-kg space probe is motionless in space. to start moving, its main engine is fired for 5 s during which time it ejects exha

ust gases at 5000 m/s. at the end of this process it is moving at 20 m/s. the approximate mass of the ejected gas is:
Physics
1 answer:
photoshop1234 [79]1 year ago
8 0
<span>When a collision occurs, the change in momentum, is the difference in the momentums of the two objects after the collision minus the momentum of the object before the collision. From the aforementioned , we have change in momentum to be final momentum after collision - momentum before collision. But then the impulse of the fird bullet is just the change in momentum. Impulse = Ft where F = Ma. Where F= force and a is the acceleration. F = 1000* a; since a = V/t; i. e change in velocity by time. So we have 5000/5 =1000ms^2. Hence F = 1000 * 1000 = 1000000N. Then our Impulse , Ft = 1000000 * 5 = 5000000N. Momentum just before collision = 1000 * 500 = 500000. After collision we have final momentum = m * 20 where m is the mass after collsion. So we have 20M - 500000 = 5000000. 20M = 5000000-500000 = 4500000. M = 225000Kg.</span>
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8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

7 0
2 years ago
A rod 16.0 cm long is uniformly charged and has a total charge of -25.0 µC. Determine the magnitude and direction of the electri
Vlada [557]

Answer:

-1.4x10^6N/C

Explanation:

Pls see attached file

8 0
2 years ago
Anthony and Maelynn are watching a football game outside on a sunny day. Anthony is wearing a black shirt and Maelynn is wearing
melomori [17]

Answer: Anthony will be warmer after the game.

Explanation :

Anthony and Maelynn are watching a football game outside on a sunny day. Anthony is wearing a black shirt and Maelynn is wearing a white shirt. Anthony will be warmer after the game. The black color is a good absorber of radiation and a bad reflector.

The black color absorbs heat until a thermal equilibrium is attained. So, it is advisable to wear cotton clothes in summers not dark colored clothes.

7 0
2 years ago
Read 2 more answers
A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°
Genrish500 [490]

Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume V_{1} = 2.00\ L

Pressure P_{1}= 1.00\ atm

Pressure P_{2}= 70.0\ kPa

Temperature T_{1}= 20.0°C = 293\ K

Temperature T_{2}= 7.00°C = 280\ K

We need to calculate the volume at mountains

Using  gas law

\dfrac{PV}{T}=\ Constant

For both temperature,

\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}

Put the value into the formula

\dfrac{101.325\times2}{293}=\dfrac{70\times V_{2}}{280}

V_{2}=\dfrac{101.325\times2\times280}{293\times70}

V_{2}=2.766\ litre

Hence, The volume at mountains is 2.766 L.

5 0
1 year ago
Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless sur
d1i1m1o1n [39]

Answer:

The speed of bullet and wooden bock coupled together, V = 22.22 m/s

Explanation:

Given that,

Mass of the bullet, m = 0.04 Kg

Mass of the wooden block, M = 0.5 Kg

The initial velocity of the bullet, u = 300 m/s

The initial velocity of the wooden block, U = 0 m/s

The final velocity of the bullet and wooden bock coupled together, V = 0 m/s

According to the conservation of linear momentum, the total momentum of the body after impact is equal to the total momentum before impact.

Therefore,

                              mV + MV = mu + MU

                               V(m+M) = mu

                                 V = mu/(m+M)

Substituting the values in the above equation,

                                V = 0.04 Kg x 300 m/s  / (0.04 Kg+ 0.5 Kg)

                                    = 22.22 m/s

Hence, the speed of bullet and wooden bock coupled together, V = 22.22 m/s

8 0
1 year ago
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