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katrin2010 [14]
2 years ago
6

A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positi

ve direction, what is the truck's resultant displacement
Physics
1 answer:
asambeis [7]2 years ago
6 0

Answer:

140m east

Explanation:

If East is positive then lets rephrase the problem into integers

A truck moves +70 m, then moves -120m, and finally moves +90m.

So totally Displacement = +70-120+90= +140m

Since east is positive, the trucks resultant displacement is 140 m east of origin

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The magnetic field 0.02 m from a wire is 0.1 t. what is the magnitude of the magnetic field 0.01 m from the same wire?
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Answer:

0.2 T

Explanation:

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3 years ago
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The electric force between two charges A. increases with distance between the charges B. increases if either one of charges gets
beks73 [17]

Answer:

Option (A) , (b) and (d) are correct option

Explanation:

According to Coulomb's law electric force between two charges is given by

F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}

From the relation we can say that force is directly proportional to magnitude of charges and inversely proportional to distance between them '

So if we increase the distance then force will decrease

Increase if any of the charge get larger

If force is attractive then both the charge will be of different sign and is force is repulsive then both the charges of same sign

From above conclusion we can say that (a), (b) and (d) are correct option

6 0
3 years ago
A string that is under 54.0 N of tension has linear density 5.20 g/m . A sinusoidal wave with amplitude 2.50 cm and wavelength 1
kicyunya [14]

Answer:

8.89288275 m/s

Explanation:

F = Tension = 54 N

\mu = Linear density of string = 5.2 g/m

A = Amplitude = 2.5 cm

Wave velocity is given by

v=\sqrt{\frac{F}{\mu}}\\\Rightarrow v=\sqrt{\frac{54}{5.2\times 10^{-3}}}\\\Rightarrow v=101.90493\ m/s

Frequency is given by

f=\frac{v}{\lambda}\\\Rightarrow f=\frac{101.90493}{1.8}\\\Rightarrow f=56.61385\ Hz

Angular frequency is given by

\omega=2\pi f\\\Rightarrow \omega=2\pi 56.61385\\\Rightarrow \omega=355.71531\ rad/s

Maximum velocity of a particle is given by

v_m=A\omega\\\Rightarrow v_m=0.025\times 355.71531\\\Rightarrow v_m=8.89288275\ m/s

The maximum velocity of a particle on the string is 8.89288275 m/s

5 0
3 years ago
If I rub a polythene piece with a wool , is there a transfer of mass from wool to polythene?​
kirill [66]

Yes, As a result, wool is positively charged while polythene is negatively charged. As a result, 1.87 1012 electrons have been transported from wool to polythene. As a result, only a sliver of mass is transferred from wool to polythene.

8 0
3 years ago
. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa radius 2cm. a. What is the speed of the water on the
lana66690 [7]

Answer:

Outlet Velocity = 192 m/s

Outlet Pressure = 510 kPa

Explanation:

Givens:

Inlet Velocity, V₁ = 12 m/s

Inlet Pressure, P₁ = 600 kPa = 600,000 Pa

Inlet Radius r₁ = 0.5 cm

Outlet Velocity, V₂ = not given (we are asked to find this)

Outlet Pressure,  P₂ = not given (we are asked to find this)

Outlet Radius, r₂ = 0.5 cm

From these, we can find the following:

Inlet Area, A₁ = π (r₁)² = π(2)² = 4π cm²

Outlet Area, A₂ = π (r₂)² = π(0.5)² = 0.25π cm²

<u>Part A :</u>

Assuming that water is incompressible, we can reason that within the same given time, the amount of volume of water entering the inlet must equal the volume of water exiting the outlet. Hence by the continuity equation (i.e. conservation of mass)

Inlet Volume flow rate = Outlet Volume flow rate

(recall that Volume flow rate in a pipe is given by Velocity x Cross Section Area), Hence the equation becomes

V₁ x A₁ = V₂ x A₂  (substituting the values that we know from above)

12 x 4π = V₂ x 0.25π  (we don't have to change all to SI units because the conversion factors on the left will cancel out the conversion factors on the right).

V₂ = (12 x 4π) / (0.25π)

V₂ = 192 m/s  (Answer)

<u>Part B:</u>

For Part B, if we assume a closed ideal system (control volume method), we can simply apply the energy equation (i.e Bernoulli's equation)

P₁ + (1/2)ρV₁ + ρgh₁ = P₂ + (1/2)ρV₂ + ρgh₂

Because the pipe is horizontal, there is no difference between h₁ and h₂, hence we can neglet this term:

P₁ + (1/2)ρV₁ = P₂ + (1/2)ρV₂  (rearranging)

P₂ = P₁ + (1/2)ρV₁ - (1/2)ρV₂

= P₁  + (1/2)ρ (V₁-V₂)

Assuming that the density of water is approx, ρ = 1000 kg/m³

P₂ = 600,000  + (1/2)(1000) (12-192)

= 600,000  + ( -90,000)

= 510,000 Pa

= 510 kPa (Answer)

8 0
3 years ago
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