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V125BC [204]
3 years ago
5

A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the

roadrunner using a net dropped from an overpass that is 10m high, how much time before the roadrunner is under the overpass, should the coyote drop the net? How far away from the overpass is the roadrunner when the coyote drops the net?
Physics
1 answer:
andreyandreev [35.5K]3 years ago
5 0

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and  how far away from the overpass is the roadrunner when the coyote drops the net.

d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

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An astronaut who is repairing the outside of her spaceship accidentally pushes away a 99.1 cm long steel rod, which flies off at
iren [92.7K]

Answer:

83.847519 mV

Explanation:

B = Magnetic field strength = 7.11 mT

L = Length of steel rod = 99.1 cm

v = Velocity of steel rod = 11.9 m/s

The induced electro motive force is given by

E=BLV\\\Rightarrow E=7.11\times 10^{-3}\times 0.991\times 11.9\\\Rightarrow E=0.083847519\ V=83.847519\ mV

The magnitude of the EMF induced between the ends of the rod is 83.847519 mV

4 0
3 years ago
Another droplet of the same mass falls 8.4 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by th
Andrej [43]

Answer:

The charge carried by the droplet is 1.330\times10^{-19}\ C

Explanation:

Given that,

Distance =8.4 cm

Time = 0.250 s

Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude 5.92\times10^4\ N/C points straight down and if the mass of the droplet is 2.93\times10^{-15} kg

We need to calculate the acceleration

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value into the formula

8.4\times10^{-2}=0+\dfrac{1}{2}\times a\times(0.250)^2

a=\dfrac{8.4\times10^{-2}\times2}{(0.250)^2}

a=2.688\ m/s^2

We need to calculate the charge carried by the droplet

Using formula of electric filed

E=\dfrac{F}{q}

q=\dfrac{ma}{E}

Put the value into the formula

q=\dfrac{2.93\times10^{-15}\times2.688}{5.92\times10^4}

q=1.330\times10^{-19}\ C

Hence, The charge carried by the droplet is 1.330\times10^{-19}\ C

8 0
3 years ago
Physics multiple choice
Aleksandr-060686 [28]

10. b

11.d

12. b

13.e

I hope this helps

3 0
4 years ago
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Solar cells are often coated with a transparent, thin film of silicon monoxide (n = 1.45) to minimize reflective losses from the
Alinara [238K]

Answer:

94.82 nm

Explanation:

We have given that wavelength \lambda=550\ nm

We have to find the minimum film thickness that produces the least reflection

minimum thickness is given by t=\frac{\lambda }{4n}

here n is given n=1.45

so minimum thickness t=\frac{550 }{4\times 1.45}=94.82\ nm

so the minimum film thickness will be 94.82 nm

7 0
4 years ago
Why is salt water a better conductor of electricity than river water?
Tamiku [17]

Answer: The conductivity of water depends on the concentration of dissolved ions in solution. ... This is because the Sodium Chloride salt dissociates into ions. Hence sea water is about a million times more conductive than fresh water.

6 0
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