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2 years ago

A 1380 kg car starts from rest at the

Physics

1 answer:

Ulleksa [173]2 years ago

6 0

Answer: 72200

Explanation:

First you must find the height for this is on an inclined hill using:

h=Lsin(angle) —> 28.0sin(11.0) = 5.34

Now you would just use the PE equation (mgh) because you are finding ME and when you starting from the top KE=0, showing that what ever answer you get from PE would equal the same for ME.

Using mgh:

m=1380

g=9.80

h=5.34

(1380)(9.8)(5.34)

=72218.16

*Rounding to the 3rd=72200

Hope this helps :)

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3 years ago

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Mercury is characterized by

cupoosta [38]

<span>It's close to the sun without much atmosphere, so it's characterized by </span><span>very extreme temperatures.

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3 years ago

A nonconducting container filled with 25 kg of water at 23°C is fitted with a stirrer, which is made to turn by gravity acting o

Paul [167]

Explanation:

Given that,

Weight of water = 25 kg

Temperature = 23°C

Weight of mass = 32 kg

Distance = 5 m

(a). We need to calculate the amount of work done on the water

Using formula of work done

$W=mgh$

$W=32\times9.8\times5$

$W=1568\ J$

The amount of work done on the water is 1568 J.

(b). We need to calculate the internal-energy change of the water

Using formula of internal energy

The change in internal energy of the water equal to the amount of the work done on the water.

$\Delta U=W$

$\Delta U=1568\ J$

The change in internal energy is 1568 J.

(c). We need to calculate the final temperature of the water

Using formula of the change internal energy

$\Delta U=mc_{p}\Delta T$

$\Delta U=mc_{p}(T_{2}-T_{1})$

$T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}$

$T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}$

$T_{2}=23.01^{\circ}\ C$

The final temperature of the water is 23.01°C.

(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.

The amount of heat is 1568 J.

Hence, This is the required solution.

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3 years ago

Why it is important to have exact standards of measurement

Tamiku [17]

Answer:

reference against doubt

Explanation:

exact standards of measurement are a reference point for situations in doubt , a line against which to test and observe.

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3 years ago

A projectile is shot straight up from the earth's surface at a speed of 10,000 km/hr. how high does it go?

mezya [45]

Naturally we assume that 10000 km/hr is initial velocity (same as being shot from a cannon), and no air resistance. With so high a velocity, the effect of diminishing gravity with increasing radius must be taken into account, so you use an energy solution. M is earth mass, r is earth radius.
KE/m = (9000000/3600)^2/2 = 3858025 J/kg
ΔPE/m = (PE(at height) - PE(at surface))/m = -GM/(r+h) + GM/r
KE/m = ΔPE/m
KE/m - GM/r = -GM/(r+h)
h = -GM / (KE/m - GM/r) - r = 335665.44 m
(Using G = 6.673E-11 Nm^2/kg^2, M = 5.9742E24 kg, r = 6378100 m)

7 0

3 years ago