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Blizzard [7]
3 years ago
14

A 10-kilogram mass is sliding along a frictionless floor with an acceleration of 5 meters per second squared. What is the magnit

ude of the net force acting on the mass?
Physics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

50N

Explanation:

Since the floor is frictionless, the net force = mass × acceleration.

given that,

mass = 10kg

and acceleration is 5 m/s²

Thus,

Force = 10 × 5

Force = 50 N

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The formation of new volcanoes and the eruption of some existing ones already. 
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A swimming pool measures 5.0 m long x 4.0 m wide x 3.0 mdeep.
QveST [7]

Answer:

a) 588,000 N

b) 294000 N

Explanation:

Given that

Density of water = 1000kg/m3

(g) = 9.8m/s2

volume is given as (V)= 5m*4m*3m

a) force will be equal to weight of water

W = mg  = \rho g V = 1000\times 9.8 \times (5*4*3) = 588,000 kg m/s^2

b) at either end

P = \frac{F}{A} = P_o + \rho gh

dF = PdA

dF = \rho g w \int h dh

F = \rho g w \frac{h^2}{2}

F = \rho g A \frac{h}{2}           [A = wh]

F = 1000\times 9.8 \times 5\times 4 \times\frac{3}{2}

F = 294000 N

3 0
4 years ago
Examine the scenario. Object A has 5 protons and 5 electrons. Object B has 5 protons and 7 electrons. Which option most accurate
Travka [436]
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Object A has a net charge of 0 because the positive and negative
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4 0
3 years ago
Which type of mirror produces images that are always upright and at the same distance from the mirror as the object is?
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7 0
4 years ago
Read 2 more answers
Two coils close to each other have a mutual inductance of 32 mH. If the current in one coil decays according to I=I0e−αt, where
fiasKO [112]

The emf induced in the second coil is given by:

V = -M(di/dt)

V = emf, M = mutual indutance, di/dt = change of current in the first coil over time

The current in the first coil is given by:

i = i₀e^{-at}

i₀ = 5.0A, a = 2.0×10³s⁻¹

i = 5.0e^(-2.0×10³t)

Calculate di/dt by differentiating i with respect to t.

di/dt = -1.0×10⁴e^(-2.0×10³t)

Calculate a general formula for V. Givens:

M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)

Plug in and solve for V:

V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))

V = 320e^(-2.0×10³t)

We want to find the induced emf right after the current starts to decay. Plug in t = 0s:

V = 320e^(-2.0×10³(0))

V = 320e^0

V = 320 volts

We want to find the induced emf at t = 1.0×10⁻³s:

V = 320e^(-2.0×10³(1.0×10⁻³))

V = 43 volts

3 0
3 years ago
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