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4 years ago

18. The displacement of an object moving 330 km North for 2 hours and an additional 220

Physics

1 answer:

motikmotik4 years ago

3 0

Answer:

Explanation:

1) Displacement of the object will be gotten using simply adding the distances since they are goinf in the same direction (positive y direction)

Displacement= 330km + 220km

Displacement = 550km North

2) For a falling body, the body will possess positive acceleration due to gravity because it is falling under the influence of gravitational force. If a body is not under the influence of gravity and is thrown up, such body wont come back down. Hence the value of acceleration due to gravity of a body falling to the ground is +9.8 m/s² (note that the unit of acceleration is m/s²)

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You push a 50 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. The coefficient of kinetic friction is 0.15. N

WARRIOR [948]

Answer:

0.68 seconds

Explanation:

Data provided in the question:

Mass of the box = 50 kg

Speed of the box = 1.0 m/s

Coefficient of friction, μ = 0.15

Now,

Force applied = μmg

Here,

g is the acceleration due to gravity = 9.8 m/s²

Thus,

F = 0.15 × 50 × 9.8

= 73.5 N

Also,

Force = Mass × Acceleration

thus,

73.5 N = 50 × a

a = 1.47 m/s²

After doubling the speed

Final speed = 2 × Initial speed

= 2 × 1 m/s

= 2 m/s

Also,

Acceleration = [change in speed] ÷ Time

1.47 = [ 2 - 1 ] ÷ Time

Time = 1 ÷ 1.47

Time = 0.68 seconds

6 0

4 years ago

A cylinder with a piston contains 0.200 mol of nitrogen at 1.50×105 Pa and 320 K . The nitrogen may be treated as an ideal gas.

Alja [10]

Answer:

$Q = -105 J$

Also we know that for cyclic process change in internal energy is always ZERO

Explanation:

First gas is compressed isobarically such that its volume is half of initial volume

So its temperature is also half

So heat given in this process is given as

$Q = nC_p \Delta T$

for diatomic gas we have

$C_p = \frac{7}{2} R$

so we will have

$Q = 0.200(\frac{7}{2}R)(160 - 320)$

$Q = -930.7 J$

Now in adiabatic process heat is not transferred

so in this process

$Q = 0$

so we have

$T_1V_1^{1.4-1} = T_2V_2^{1.4-1}$

$(160)(\frac{V}{2})^{0.4} = T_2(V)^{0.4}$

$T_2 = 121.26 K$

Now it is again reached to original pressure

so temperature will become initial temperature

so heat given in that part

$Q_3 = nC_v\Delta T$

here we know that

$C_v = \frac{5}{2}R$

$Q_3 = (0.200)(\frac{5}{2}R)(320 - 121.26)$

$Q_3 = 825.76 J$

So total heat given to the system is

$Q = -930.7 + 0 + 825.76$

$Q = -105 J$

Also we know that for cyclic process change in internal energy is always ZERO

3 0

3 years ago

A magnetic field is aligned perpendicular to the plane of a circular loop of wire with radius 5.0 cm and 20 turns. The magnetic

Mariana [72]

Answer:

$V = 0.157(-36 e^{-3t} + 1.5)$

$t = 1.24 s$

Part b)

$EMF = 0.136(-36e^{-3t} + 1.5)$

Explanation:

magnetic field due to external source is given as

$B = 12 e^{-3t} + 1.5 t + 6$

area of the loop is given as

$A = \pi r^2$

$A = \pi(0.05)^2$

$A = 7.85 \times 10^{-3} m^2$

Now we have

$\phi = NBAcos0$

$\phi = (20)(12e^{-3t} + 1.5t + 6)(7.85 \times 10^{-3})$

$V = \frac{d\phi}{dt}$

$V = 0.157\frac{d}{dt}(12e^{-3t} + 1.5t + 6)$

$V = 0.157(-36 e^{-3t} + 1.5)$

now we need to find the time at which voltage is 0.1 Volts so we have

$0.1 V = 0.157(-36e^{-3t} + 1.5)$

$t = 1.24 s$

Part b)

If magnetic field is inclined at an angle of 30 degree with the normal of the loop then

$\phi = NBAcos30$

now we know that induced EMF is given as

$EMF = \frac{d\phi}{dt}$

$EMF = NAcos30\frac{dB}{dt}$

$EMF = (20)(7.85 \times 10^{-3})cos30(\frac{d}{dt}(12e^{-3t} + 1.5t + 6))$

$EMF = 0.136(-36e^{-3t} + 1.5)$

7 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS AND I NEED ALL CORRECT ANSWERS ONLY!!!

ryzh [129]

iron, cobalt and Nickel is the answer

6 0

3 years ago

A car is being lifted by a crane so that it moves upwards at a constant speed.

anygoal [31]

Answer:

Explanation:

Work done is defined as the product of force and the displacement in the direction of force.

W = F x d x Cos θ

Where, θ be the angle between force and the displacement.

The force of gravity acts downwards and the displacement is in upward, so the angle is 180. cos 180 = -1 , so the work done by the force of gravity is negative.

The force of rope acting in upwards direction and displacement also in upwards direction.

So, the work done is positive

7 0

4 years ago