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Nikitich [7]
3 years ago
5

Two rocks weighing 5 Newtons and 10 Newtons are dropped simultaneously from the same height onto a Coyote. After three seconds o

f free fall (neglecting air resistance), compared to the 5 N rock, the 10 N rock has greater
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
5 0

<h2>Potential energy lost by 10 N rock will be greater</h2>

Explanation:

Two rocks of 5N and 10N falls from the same height . Thus they will loose the potential energy.

The potential energy lost = mass x acceleration due to gravity x height

The potential energy lost by first 5 N rock = 5 h

Because weight of rock m g = 5 N

Similarly , the potential energy lost by 10 N Rock = 10 h

here weight of rock m g = 10 N

Thus comparing these two , the potential energy lost by 10 N rock is greater than that of 5 N rock .

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starting from a stop a traffic signal, a car speeds up to 20 m/s in 5 seconds. calculate the acceleration of the car.
Ira Lisetskai [31]

Answer:

5.867 m/s^2

Explanation:

Initial Speed > 0 m/s

Final Speed > 20 m/s

Time > 5 sec.

8 0
3 years ago
Which variety of moth, C or Y, is more likely to be killed by insect eating birds?
Svetlanka [38]
Moth X is more likely to be eaten by insect killing birds first. This is because Moth Y blends in with the tree trunk more and is hidden from birds. Its wings camouflage with the tree trunk, hiding it from sight. 
7 0
3 years ago
A camera is equipped with a lens with a focal length of 34 cm. When an object 2.4 m (240 cm) away is being photographed, what is
puteri [66]

Answer:

The magnification is -6.05.

Explanation:

Given that,

Focal length = 34 cm

Distance of the image =2.4 m = 240 cm

We need to calculate the distance of the object

\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}

Where, u = distance of the object

v = distance of the image

f = focal length

Put the value into the formula

\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}

\dfrac{1}{u}=\dfrac{103}{4080}

u =\dfrac{4080}{103}

The magnification is

m = \dfrac{-v}{u}

m=\dfrac{-240\times103}{4080}

m = -6.05

Hence, The magnification is -6.05.

6 0
3 years ago
Find the total electric charge of 1.7 kg of electrons. me=9.11×10−31kg, e=1.60×10−19C.
Gelneren [198K]

Answer:

2.99\cdot 10^{11}C

Explanation:

The mass of one electron is

m_e = 9.11\cdot 10^{-31}kg

So the number of electrons contained in M=1.7 kg of mass is

N=\frac{M}{m_e}=\frac{1.7 kg}{9.11\cdot 10^{-31}kg}=1.87\cdot 10^{30}

The charge of one electron is

e=1.60\cdot 10^{-19} C

So, the total charge of these electrons is equal to the charge of one electron times the number of electrons:

Q=Ne=(1.87\cdot 10^{30})(1.6\cdot 10^{-19}C)=2.99\cdot 10^{11}C

8 0
3 years ago
An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b
Mariana [72]

Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

6 0
3 years ago
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