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Andru [333]
2 years ago
9

11. A plow pushes 100 kg of snow with 300 N of force. How much is the pile of snow accelerated?

Physics
1 answer:
Fed [463]2 years ago
3 0

Answer:

Explanation:

This is an application of Newton's second Law.

Formula

F = m * a

F = 300 N

m = 100 kg

a = ?

F = m * a

300N = 100 kg * a            Divide by 100

300N/100kg = a

a = 3 m/sec^2

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A small ferryboat is 4.70 m wide and 6.10 m long. When a loaded truck pulls onto it, the boat sinks an additional 5.00 cm into t
geniusboy [140]

Answer:

   M = 1433.5 kg

Explanation:

This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,

              B = ρ g V

with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition

           Σ F = 0

           B-W = 0

           B = W

       body weight

           W = M g

the volume is

           V = l to h

           rho_liquid g (l to h) = M g

           M = rho_liquid l a h

           

we calculate

            M = 1000 4.7 6.10 0.05

           M = 1433.5 kg

6 0
3 years ago
How can hypotheses best be tested
Lerok [7]

In order for a hypothesis to be tested, and experiment needs to be designed.

Many trials need to be runned in order to get accurate results and to form a valid conclusion that can prove or disprove the hypothesis

3 0
3 years ago
Y=mx+b if y=8.18 m=1.31 b=17.2 then find x
Dafna1 [17]

Plug in the corresponding values into y = mx + b

8.18 in for y

1.31 in for m

17.2 in for b

8.18 = 1.31x + 17.2

Now bring 17.2 to the left side by subtracting 17.2 to both sides (what you do on one side you must do to the other). Since 17.2 is being added on the right side, subtraction (the opposite of addition) will cancel it out (make it zero) from the right side and bring it over to the left side.

8.18 - 17.2 = 1.31x

-9.02 = 1.31x

Then divide 1.31 to both sides to isolate x. Since 1.31 is being multiplied by x, division (the opposite of multiplication) will cancel 1.31 out (in this case it will make 1.31 one) from the right side and bring it over to the left side.

-9.02/1.31 = 1.31x/1.31

x ≈ -6.8855

x is roughly -6.89

Hope this helped!

~Just a girl in love with Shawn Mendes

8 0
3 years ago
Read 2 more answers
A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of
Gekata [30.6K]

Answer:

a) P=2450\ Pa

b) \delta h=23.162\ cm

Explanation:

Given:

height of water in one arm of the u-tube, h_w=25\ cm=0.25\ m

a)

Gauge pressure at the water-mercury interface,:

P=\rho_w.g.h_w

we've the density of the water =1000\ kg.m^{-3}

P=1000\times 9.8\times 0.25

P=2450\ Pa

b)

Now the same pressure is balanced by the mercury column in the other arm of the tube:

\rho_w.g.h_w=\rho_m.g.h_m

1000\times 9.8\times 0.25=13600\times 9.8\times h_m

h_m=0.01838\ m=1.838\ cm

<u>Now the difference in the column is :</u>

\delta h=h_w-h_m

\delta h=25-1.838

\delta h=23.162\ cm

7 0
2 years ago
A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t
Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=\sqrt{\frac{Gmm}{R+h} }         [(R+h) is the distance of the satellite   from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = h_{i} = 98 km = 98000 m

∴Initial Energy (E_{i})  = \frac{1}{2}mv^{2} + \frac{-GMm}{(R+h_{i} )}

Substituting v=\sqrt{\frac{GMm}{R+h_{i} } } in the above equation and simplifying we get,

E_{i} = \frac{-GMm}{2(R+h_{i}) }

Similarly for final condition,

h=h_{f} = 198km = 198000 m

∴Final Energy(E_{f}) = \frac{-GMm}{2(R+h_{f}) }

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = E_{f} - E_{i}

        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

Substituting ,

M = 6 × 10^{24} kg

m = 1036 kg

G = 6.67 × 10^{-11}

R = 6400000 m

h_{i} = 98000 m

h_{f} = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = \frac{1}{2}m[v_{f} ^{2} - v_{i} ^{2}]

                                                          = \frac{GMm}{2}[\frac{1} {R+h_{f} } - \frac{1} {R+h_{i} }]

                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} }]

                                                             = 2ΔE

                                                             = 968.907 MJ

3 0
3 years ago
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