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kondor19780726 [428]
3 years ago
11

A bird can fly 25 km/h. How long does it take to fly 3.5km?

Physics
1 answer:
monitta3 years ago
6 0

The equation of motion that we can use in this case is:

t = d / v

where t is time, d is distance, v is velocity

 

Therefore calculating for t:

t = 3.5 km / (25 km / h)

<span>t = 0.14 h = 8.4 minutes</span>

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The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant
Natasha2012 [34]

Answer:

D) Q/2

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Q=CV (1)

In the first part of the problem, we have that the charge stored on the capacitor is Q, when the voltage supplied is V. The capacitance of the parallel-plate capacitor is given by

C=\frac{\epsilon_0 A}{d}

where \epsilon_0 is the vacuum permittivity, A is the area of the plates, d is the separation between the plates.

Later, the voltage of the battery is kept constant, V, while the separation between the plates of the capacitor is doubled: d'=2d. The capacitance becomes

C'=\frac{\epsilon_0 A}{d'}=\frac{\epsilon_0 A}{2d}=\frac{C}{2}

And therefore, the new charge stored on the capacitor will be

Q'=C'V=\frac{C}{2}V=\frac{Q}{2}

6 0
3 years ago
A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
4 years ago
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