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1 year ago

A 60.0 kg object is moving east at 8.00 m/s, and then slows down to 4.00 m/s. How much work was done?

Physics

2 answers:

Artist 52 [7]1 year ago

8 0

The answer is -1440. That is letter A.

almond37 [142]1 year ago

3 0

Answer:

Work done, W = -1440 J

Explanation:

It is given that,

Mas of an object, m = 60 kg

Initial velocity of the object, u = 8 m/s

Final velocity of an object, v = 4 m/s

We know that work done is equal to change in kinetic energy of an object as per work energy theorem.

$W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

$W=\dfrac{1}{2}m(v^2-u^2)$

Putting all the values in above equation we get :

$W=\dfrac{1}{2}\times 60\ kg((4\ m/s)^2-(8\ m/s)^2)$

W = -1440 J

So, the work done by the object is -1440 Joules. Hence, the correct option is (A) " -1440 J ".

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he first excited state of the helium atom lies at an energy 19.82 eV above the ground state. If this excited state is three-fold

bekas [8.4K]

Answer:

Relative population is 2.94 x 10⁻¹⁰.

Explanation:

Let N₁ and N₂ be the number of atoms at ground and first excited state of helium respectively and E₁ and E₂ be the ground and first excited state energy of helium respectively.

The ratio of population of atoms as a function of energy and temperature is known as Boltzmann Equation. The equation is:

$\frac{N_{1} }{N_{2} }$ = $\frac{g_{1}e^{\frac{-E_{1} }{KT} } }{g_{2}e^{\frac{-E_{2} }{KT} }}$

$\frac{N_{1} }{N_{2} }$ = $\frac{g_{1}e^{\frac{-(E_{1}-E_{2}) }{KT} } }{g_{2}}$

Here g₁ and g₂ be the degeneracy at two levels, K is Boltzmann constant and T is equilibrium temperature.

Put 1 for g₁, 3 for g₂, -19.82 ev for (E₁ - E₂) and 8.6x10⁵ ev/K for K and 10000 k for T in the above equation.

$\frac{N_{1} }{N_{2} }$ = $\frac{1\times e^{\frac{-(-19.82)}{8.6\times 10^{-5}\times 10000} } }{3}$

$\frac{N_{1} }{N_{2} }$ = 3.4 x 10⁹

$\frac{N_{2} }{N_{1} }$ = 2.94 x 10⁻¹⁰

5 0

2 years ago

At one point in the rescue operation, breakdown vehicle A is exerting a force of 4000 N and breakdown vehicle B is exerting a fo

lukranit [14]

Answer:

1.) Magnitude = 5596 N

2.) Direction = 60 degrees

Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N

Let us resolve the two forces into X and Y component

Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N

Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )

= 2000 + 2828.43

= 4828.43 N

The resultant force R will be

R = sqrt ( X^2 + Y^2 )

Substitutes the forces at X component and Y component into the formula

R = sqrt ( 2828.43^2 + 4828.43^2 )

R = sqrt ( 31313752.53 )

R = 5595.87 N

The direction will be

Tan Ø = Y/X

Substitute Y and X into the formula

Tan Ø = 4828.43 / 2828.43

Tan Ø = 1.707106

Ø = tan^-1( 1.707106 )

Ø = 59.64 degree

Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.

8 0

2 years ago

The current in a long solenoid of radius 6 cm and 17 turns/cm is varied with time at a rate of 5 A/s. A circular loop of wire of

jonny [76]

Answer:

53.63 μA

Explanation:

radius of solenoid, r = 6 cm

Area of solenoid = 3.14 x 6 x 6 = 113.04 cm^2 = 0.0113 m^2

n = 17 turns / cm = 1700 /m

di / dt = 5 A/s

The magnetic field due to the solenoid is given by

B = μ0 n i

dB / dt = μ0 n di / dt

The rate of change in magnetic flux linked with the solenoid =

Area of coil x dB/dt

= 3.14 x 8 x 8 x 10^-4 x μ0 n di / dt

= 3.14 x 64 x 10^-4 x 4 x 3.14 x 10^-7 x 1700 x 5 = 2.145 x 10^-4

The induced emf is given by the rate of change in magnetic flux linked with the coil.

e = 2.145 x 10^-4 V

i = e / R = 2.145 x 10^-4 / 4 = 5.36 x 10^-5 A = 53.63 μA

6 0

1 year ago

An inventive child named Nick wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rop

Oduvanchick [21]

UHHH WHAT? I DONT GET THAT AT ALLOW

5 0

2 years ago

A box of mass 3kg is lifted 1.5m onto a shelf. Calculate the change in its gravitational potential energy. The gravitational fie

Vedmedyk [2.9K]

Answer:

The change in gravitational potential energy is 45 J.

Explanation:

Given that,

Mass = 3 kg

Distance = 1.5 m

Gravitational field strength = 10 N/kg

We need to calculate the change in gravitational potential energy

Using formula of gravitational potential energy

$Change\ in\ gravitational\ potential\ energy =gravitational\ field\ strength\times mass\times distance$

Put the value into the formula

$Change\ in\ gravitational\ potential\ energy =10\times3\times1.5$

$Change\ in\ gravitational\ potential\ energy =45\ J$

Hence, The change in gravitational potential energy is 45 J.

5 0

2 years ago