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ludmilkaskok [199]

3 years ago

5

A 30.0-kgkg box is being pulled across a carpeted floor by a horizontal force of 230 NN , against a friction force of 210 NN . W

hat is the acceleration of the box?

1 answer:

Agata [3.3K]3 years ago

4 0

Answer:

The acceleration of the box is 0.67 m/s²

Explanation:

Given that,

Mass of box = 30.0 kg

Horizontal force = 230 N

Friction force = 210 N

We need to calculate the acceleration of the box

Using balance equation

$F-f_{k}=ma$

$a=\dfrac{F-f_{k}}{m}$

Where, F = horizontal force

$f_{k}$ =frictional force

m= mass of box

a = acceleration

Put the value into the formula

$a=\dfrac{230-210}{30}$

$a=0.67\ m/s^2$

Hence, The acceleration of the box is 0.67 m/s²

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A 25,000-kg train car moving at 2.50 m/s collides with and connects to a train car of equal mass moving in the same direction at

Margarita [4]

Answer:

14062.5 J

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

V = (m₁u₁ + m₂u₂)/(m₁+m₂).................1

Where V = common velocity after collision

Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

Substitute into equation 1

V = [25000(2.5) + 25000(1)]/(25000+25000)

V = (62500+25000)/50000

V = 87500/50000

V = 1.75 m/s.

Note: The collision is an inelastic collision as such there is lost in kinetic energy of the system.

Total Kinetic energy before collision = kinetic energy of the first train car + kinetic energy of the second train car

E₁ = 1/2m₁u₁² + 1/2m₂u₂²........................ Equation 2

Where E₁ = Total kinetic energy of the body before collision, m₁ and m₂ = mass of the first train car and second train car respectively. u₁ and u₂ = initial velocity of the first train car and second train car respectively.

Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

Substitute into equation 2

E₁ = 1/2(25000)(2.5)² + 1/2(25000)(1.0)²

E₁ = 12500(6.25) + 12500

E₁ = 78125+12500

E₁ = 90625 J.

Also

E₂ = 1/2V²(m₁+m₂)....................... Equation 3

Where E₂ = total kinetic energy of the system after collision, V = common velocity, m₁ and m₂ = mass of the first and second train car respectively.

Given: V = 1.75 m/s, m₁ = m₂ = 25000 kg

Substitute into equation 3

E₂ = 1/2(1.75)²(25000+25000)

E₂ = 1/2(3.0625)(50000)

E₂ = (3.0625)(25000)

E₂ = 76562.5 J.

Lost in kinetic Energy of the system = E₁ - E₂ = 90625 - 76562.5

Lost in kinetic energy of the system = 14062.5 J

5 0

3 years ago

Three Carnot engines operate between the following temperature limits.

Kipish [7]

Answer:c>a>b

Explanation:

Given

All the engine extracts same amount of heat(Q) from High-temperature reservoir

For a) 400 and 500 K

$\eta _{engine}=1-\frac{T_L}{T_H}$

$\eta _{engine}=\frac{work\ supplied}{heat\ supplied}$

$1-\frac{400}{500}=\frac{W_a}{Q}$

$W_a=\frac{Q}{5}$

For b)500 K and 600K

$1-\frac{500}{600}=\frac{W_b}{Q}$

$W-b=\frac{Q}{6}$

For c) 400 K and 600 K

$1-\frac{400}{600}=\frac{W_c}{Q}$

$W_c=\frac{2Q}{3}$

So c will give the highest amount of work

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6 0

4 years ago

Light that has a wavelength of 600 nm strikes a metal surface, and a stream of electrons is ejected from the surface. If light o

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Answer:

The maximum kinetic energy of the electrons emitted from the surface will be greater.

Option 3 is correct.

Explanation:

Given that,

Wavelength $\lambda_{1}=600\ nm$

Wavelength $\lambda_{2}=500\ nm$

We need to calculate the energy for first wavelength

Using formula of energy

$E=hf$

$E=h\dfrac{c}{\lambda}$

Put the value into the formula

$E_{1}=6.67\times10^{-34}\times\dfrac{3\times10^{8}}{600\times10^{-9}}$

$E_{1}=3.33\times10^{-19}\ J$

We need to calculate the energy for second wavelength

Put the value into the formula

$E_{2}=6.67\times10^{-34}\times\dfrac{3\times10^{8}}{500\times10^{-9}}$

$E_{2}=4.002\times10^{-19}\ J$

So, E₂>E₁

Hence, The maximum kinetic energy of the electrons emitted from the surface will be greater.

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Answer:

hope this helps

Explanation:

solid particles are arranged according to the size of the object

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while gas particles are arranged scatterly like they spread.

3 0

3 years ago

Read 2 more answers