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Aleksandr-060686 [28]
4 years ago
15

A 45.0-kg girl is standing on a 147-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,

frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.53 m/s to the right relative to the plank. (Let the direction the girl is moving in be positive. Indicate the direction with the sign of your answer.)
Physics
1 answer:
max2010maxim [7]4 years ago
6 0

Answer:

v_1 = 1.17 m/s

Explanation:

given,

mass of the girl (m₁)= 45 kg

mass of the plank(m₂) = 147 kg

constant velocity of girl relative to plank  = 1.53 m/s

velocity relative to ice berg(v₁) = ?

conservation of momentum

momentum of system = 0

m_1v_1 = m_2v_2

45 \times v_1 = 147 \times (1.53 - v_1)

45v_1 = 224.91 - 147 v_1

192 v_1 = 224.91

v_1 = \dfrac{224.91}{192}

v_1 = 1.17 m/s

hence, the velocity relative to ice berg is 1.17 m/s

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Answer:

A. The box experiences more friction on the carpeted floor

Explanation:

Friction is the force that opposes the motion of an object when it slides along a surface. The magnitude of the friction is given by

F=\mu mg

where m is the mass of the object, g is the acceleration due to gravity, and \mu is the coefficient of friction, which depends on the type of material of the surface: the larger this coefficient, the stronger the friction, the more difficult is to push the box along the surface. Generally, a smooth surface has a lower coefficient of friction, while a rough surface has a larger coefficient of friction.

In this case, Ethan find it easier to push the box on the marble floor, because marble is smoother than the carpet and so friction is weaker, while for the carpeted floor the coefficient of friction is larger and so the friction is stronger, making it more difficult to push the box.

3 0
3 years ago
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True or False. If your skin is wet, the body's resistance to electric shocks increases.
Svetllana [295]

Answer:

Pretty sure its false, water increases electric shock

Explanation:

4 0
3 years ago
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Giving a test to a group of students, the grades and gender are summarized belowGrades vs. Gender ABCMale10316Female465If one st
strojnjashka [21]

Answer:

20.45%

Explanation:

The probability that the student got a B is

\frac{total\#of\text{ students that got B}}{\text{total students }}\times100\%

Now, how many students are there in total?

The answer is

10+3+16+4+6+5=44\; \text{students}

How many students got a B?

The answer is

3+6=9\; \text{students}

therefore, the probability that the student has got a B is

\frac{9\text{ students }}{44\text{ students }}\times100\%=20.45\%

Hence, the probability that a student has got a B is 20.45%

5 0
1 year ago
The Sun delivers an average power of 1.575 W/m2 to the top of Neptune's atmosphere. Find the magnitudes of max and max for the e
FrozenT [24]

Answer:

1.1486813808\times 10^{-7}\ T

34.46 V/m

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

c = Speed of light = 3\times 10^8\ m/s

I = Intensity = 1.575 W/m²

The maximum magnetic field intensity is given by

B_m=\sqrt{\dfrac{2\mu_0I}{c}}\\\Rightarrow B_m=\sqrt{\dfrac{2\times 4\pi \times 10^{-7}\times 1.575}{3\times 10^8}}\\\Rightarrow B_m=1.1486813808\times 10^{-7}\ T

The magnetic field intensity is 1.1486813808\times 10^{-7}\ T

The maximum electric field intensity is given by

E_m=B_m\times c\\\Rightarrow E_m=1.1486813808\times 10^{-7}\times 3\times 10^8\\\Rightarrow E_m=34.46\ V/m

The  electric field intensity is 34.46 V/m

8 0
3 years ago
A transformer has a primary coil with 375 turns of wire and a secondary coil with 1,875 turns. An AC voltage source connected ac
Sonbull [250]

Answer:

The rms voltage (in V) measured across the secondary coil is 459.62 V

Explanation:

Given;

number of turns in the primary coil, Np = 375 turns

number of turns in the secondary coil, Ns = 1875 turns

peak voltage across the primary coil, Ep = 130 V

peak voltage across the secondary coil, Es = ?

\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V

The rms voltage (in V) measured across the secondary coil is calculated as;

V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V

Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V

7 0
3 years ago
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