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svet-max [94.6K]
2 years ago
8

The coefficient of kinetic friction between a suitcase and the floor is 0.272. You may want to review (Pages 196 - 203) . Part A

If the suitcase has a mass of 80.0 kg , how far can it be pushed across the level floor with 660 J of work
Physics
1 answer:
vazorg [7]2 years ago
8 0

Answer:

Explanation:

The work required to push will be equal to work done by friction . Let  d be the displacement required .

force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction

work done by force of friction

mg x μ x d   = 660

80 x 9.8 x .272 x d = 660

d = 3 .1 m .

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The magnetic field in the region between the poles of an electromagnet is uniform at any time, (1 point) but is increasing at th
olga nikolaevna [1]

Answer:

B)

The magnitude of induced emf in the conducting loop is 0.99 mV.

Explanation:

Rate of increase in magnetic field per unit time = 0.090 T/s

Area of the conducting loop = 110 cm^2 = 0.0110 m^2

Electromagnetic induction is the production of an emf or voltage in a coil of wire due to a changing magnetic field through the coil.

Induced e.m.f is given as:

EMF = (-N*change in magnetic field/time)*Area

EMF = rate of change of magnetic field per unit time * Area

EMF = 0.090 * 0.0110

EMF = 0.00099 V

EMF = 0.99 mV

5 0
3 years ago
A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe
notsponge [240]

Answer:

Q = \frac{0.068}{E}

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field

while weight of the ball is vertically downwards

so here we have

QE = F_x

mg = F_y

since string makes 30 degree angle with the vertical so we will have

tan\theta = \frac{F_x}{F_y}

tan30 = \frac{QE}{mg}

Q = \frac{mg}{E}tan30

Q = \frac{0.012\times 9.81}{E} tan30

Q = \frac{0.068}{E}

where E = electric field intensity

5 0
3 years ago
You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,
lbvjy [14]

Answer:

The value is T_c  =  12 .1 ^oC

Explanation:

From the question we are told that

The mass of the ice cube is m_i  =  7.50 *10^{-3} \  kg

The temperature of the ice cube is T_i = 0^o C

The mass of the copper cube is m_c  =  0.540 \  kg

The final temperature of both substance is T_f  =  0^oC

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

Q =  m_c  *  c_c *  [T_c  -  T_f ]

The specific heat of copper is c_c  = 385J/kg \cdot  ^oC

Generally the heat gained by the ice cube is mathematically represented as

Q_1 =  m_i * L

Here L is the latent heat of fusion of the ice with value L  =  3.34 * 10^{5} J/kg

So

Q_1 =  7.50 *10^{-3} * 3.34 * 10^{5}

=> Q_1 =  2505 \ J

So

2505  =  0.540  *  385 *  [T_c  - 0 ]

=>    T_c  =  12 .1 ^oC

4 0
2 years ago
Researcher measures the thickness of a layer of benzene (n = 1.50) floating on water by shining monochromatic light onto the fil
earnstyle [38]

Answer:

Minimum thickness; t = 9.75 x 10^(-8) m

Explanation:

We are given;

Wavelength of light;λ = 585 nm = 585 x 10^(-9)m

Refractive index of benzene;n = 1.5

Now, let's calculate the wavelength of the film;

Wavelength of film;λ_film = Wavelength of light/Refractive index of benzene

Thus; λ_film = 585 x 10^(-9)/1.5

λ_film = 39 x 10^(-8) m

Now, to find the thickness, we'll use the formula;

2t = ½m(λ_film)

Where;

t is the thickness of the film

m is an integer which we will take as 1

Thus;

2t = ½ x 1 x 39 x 10^(-8)

2t = 19.5 x 10^(-8)

Divide both sides by 2 to give;

t = 9.75 x 10^(-8) m

8 0
2 years ago
Matter and energy can convert into each other. True False
Valentin [98]

the answer is True you can convert matter and energy

8 0
3 years ago
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