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lara [203]
3 years ago
8

A particle moving uniformly along the x axis is located at 11.3m at 0.588s and at 3.38m at 4.67s. Find its displacement during t

his time interval.
Physics
2 answers:
defon3 years ago
7 0
7.92 meters, since its very clear that the particle at 4.67 seconds it was opposing the initial direction therefore you can say 11.3m - 3.387m and get your answer
Yakvenalex [24]3 years ago
5 0
Displacement = (distance between start and end points) in the direction of (direction from start to end point). Distance = (11.3-3.38)= 7.92 m. Direction = the negative 'x' direction.
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Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a m
marusya05 [52]

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

4 0
4 years ago
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7 0
4 years ago
It takes 180 kj of work to accelerate a car from 21.0 m/s to 27.0 m/s. what is the car's mass?
beks73 [17]
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4 0
3 years ago
A typical jet airliner has a cruise airspeed of 900 km/h , which is its speed relative to the air through which it is flying. If
Luba_88 [7]

Explanation:

It is given that,

Speed of the jet airplane with respect to air, v_{PA} = 900\ km/h          

If the wind at the airliner’s cruise altitude is blowing at 100 km/h from west to east, v_{AG}= 100\ km/h

(A) Let v_{PG} is the speed of the airliner relative to the ground if the airplane is flying from west to east,

v_{PG}=900-100=800\ km/h

(B) Let v'_{PG} is the speed of the airliner relative to the ground if the airplane is flying from east to west,

v'_{PG}=900+100=1000\ km/h

Hence, this is the required solution.                                            

8 0
3 years ago
Ballistic data obtained on a firing range show that aerodynamic drag reduces the speed of a .44 magnum revolver bullet from 250
m_a_m_a [10]

Answer:

0.363999909622

Explanation:

F = Force

m = Mass = 15.6 g

C = Drag coefficient

ρ = Density of air = 1.21 kg/m³

A = Surface area = \dfrac{\pi}{4}d^2

v = Terminal velocity = v=210\ m/s

s = Displacement = 150 m

a=\dfrac{v^2-u^2}{2s}

Force is given by

F = ma

F=\dfrac{1}{2}\rho CAv^2\\\Rightarrow ma=\dfrac{1}{2}\rho CAv^2\\\Rightarrow m\dfrac{v^2-u^2}{2s}=\dfrac{1}{2}\rho CAv^2\\\Rightarrow C=2\times m\dfrac{v^2-u^2}{2s}\times\dfrac{1}{\rho Av^2}\\\Rightarrow C=2\times15.6\times 10^{-3}\dfrac{210^2-250^2}{2\times 150}\times\dfrac{1}{1.21\times\dfrac{\pi}{4}\times (11.2\times 10^{-3})^2(210)^2}\\\Rightarrow C=-0.363999909622

The drag coefficient is 0.363999909622 (ignoring negative sign)

4 0
3 years ago
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