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Stels [109]
3 years ago
13

A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.

How high above the ground was the airplane when the supplies were dropped? (Hint: d = 1/2gt2; g = -9.8 m/s2)
Question 3 options:

120 meters


490 meters


980 meters


49 meters
Physics
1 answer:
Dahasolnce [82]3 years ago
3 0

The answer is B) 490.

Just fill in the formula

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The charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electro
shusha [124]

Answer:

1.6\times 10^{-18} C  

Explanation:

The fundamental charge of a single electron is 1.6\times 10^{-19} C.

If there are 10 excess electrons, the net charge that would be measured should be 10 times the fundamental charge of a single electron:

Q=nq_e\\Q= 10\times 1.6\times 10^{-19} C\\Q= 1.6 \times 10^{-18} C

3 0
3 years ago
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
3 years ago
Dos cargas Q1=2pc y Q2=4pc estan separadas por una distancia de 6cm ¿con que fuerza se atraen?
noname [10]

Here we can use coulomb's law to find the force between two charges

As per coulombs law

]tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we have

k = 9 * 10^9

q_1 = 2pC

q_2 = 4pC

r = 6cm = 0.06 m

now by using the above equation we have

F = \frac{9*10^9 * 2*10^{-12} * 4*10^{-12}}{0.06^2}

F = 2 * 10^{-11} N

so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.

3 0
3 years ago
Which word in the sentence is a gerund?
OleMash [197]
A gerund usually ends wit -ing, so we must find the word that includes that.

Loading includes ing at the end.

Final answer: b. loading
8 0
3 years ago
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A motorcycle accelerates from at a rate of 4m/s 2 while traveling 60m what it’s the motorcycles velocity at the end of this moti
fenix001 [56]

Answer: C

Explanation: 60 divided by 4 =15

5 0
3 years ago
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