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Tpy6a [65]
3 years ago
15

A 10 kg mass rests on a table. What acceleration will be generated when a force of 20 N is applied and encounters a frictional f

orce of 15 N?
Physics
1 answer:
deff fn [24]3 years ago
6 0
Net force acting on mass = 20 - 15 = 5N. ( subtracted cuz friction always opposes the motion i.e it always acts in direction opposite to the motion of the object). According to Newton's 2nd law of motion, F(net) = ma. a =F (net) / m = 5/10 = 0.5 m/s^2. Hope it helps :)
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A proton and an electron are moving in the +x direction in a magnetic field in the +z
Cloud [144]

Answer:

Explanation:

A proton and electron are moving in the positive x direction, this shows that their velocity will be in the positive x direction

V = v•i

Magnetic field Is the positive z direction

B = B•k

A. For proton.

Proton has a positive charge of q

Direction of force on proton

Force is given as

F = q(v×B)

F = q( v•i × B•k)

F = qvB (i×k)

From vectors i×k = -j

F = -qvB •j

Then, for the positive charge, the force will act in the negative direction of the y-axis

B. For electron

Electron has a negative of -q

Direction of force on proton

Force is given as

F = q(v×B)

F = -q( v•i × B•k)

F = -qvB (i×k)

From vectors i×k = -j

F = --qvB •j

F = qvB •j

Then, for the negative charge, the force will act in the positive direction of the y-axis

5 0
3 years ago
Does sound travel much faster than light ?
Kitty [74]

Answer:

nothing travels faster than light

Example:

You’ll always see lightning before you hear it, because typically lightning will be a mile away, two miles away.

4 0
3 years ago
The velocity of a wave with a wavelength of 4.700 m and frequency of 54.00 Hz is
Feliz [49]
The answer is 253.8 m/s
6 0
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The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

The ideal gas equation:

p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

20 C = 293 K

p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

Δu = Cv * Δt

Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

7 0
3 years ago
Energy flows from the sun to _______ to consumers and eventually to _______
Kay [80]

Answer:

the answer is c. producers, detrivores

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