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White raven [17]

2 years ago

14

Acceleration is always measured in meters per second squared. Why is acceleration not measured in meters per second like speed?

Explain.

1 answer:

Taya2010 [7]2 years ago

6 0

Answer:

Because 'distance per second' is a velocity, not an acceleration.

Explanation:

Because 'distance per second' is a velocity, not an acceleration. For example, at 1 m/s an object is travelling a distance of 1 metre every second. But a rate of acceleration is a steady increase in velocity. So at 1 m/s^2, an object's velocity is increasing by 1 m/s every second.

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An American traveler in China carries a transformer to convert China's standard 220 V to 120 V so that she can use some small ap

Arisa [49]

Answer:

(a) The ratio of turns in the primary and secondary coils of her transformer is 1.833

(b) The ratio of input to output current is 0.55

(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.

Explanation:

Given;

input voltage, $V_p$ = 220 V

output voltage, $V_s$ = 120 V

General transformer equation is given as;

$\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}$

where;

Np is number of turns in the primary coil

Ns is number of turns in the secondary coil

Is - is the secondary current or output current

Ip - is the primary current or input current

(a) The ratio of turns in the primary and secondary coils of her transformer;

$\frac{N_p}{N_s} = \frac{V_p}{V_s} \\\\\frac{N_p}{N_s} = \frac{220}{120} = 1.833$

(b) The ratio of input to output current;

$\frac{I_p}{I_s} = \frac{V_s}{V_p} \\\\\frac{I_p}{I_s} = \frac{120}{220} \\\\\frac{I_p}{I_s} = 0.55$

(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.

8 0

2 years ago

Find the acceleration of the blocks when the system is released. The coefficient of kinetic friction is 0.4, and the mass of eac

grin007 [14]

Answer:

a = 4.9(1 - sinθ - 0.4cosθ)

Explanation:

Really not possible without a complete setup.

I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g

F = ma

mg - mgsinθ - μmgcosθ = (m + m)a

mg(1 - sinθ - μcosθ) = 2ma

½g(1 - sinθ - μcosθ) = a

maximum acceleration is about 2.94 m/s² when θ = 0

acceleration will be zero when θ is greater than about 46.4°

8 0

3 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

3 0

3 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

2 years ago

What is the most important safety rule in cycling?

Mekhanik [1.2K]

Answer:

Wear a helmet. Stay visible; use bike lights and/or wear bright clothes. Look and Signal; use hand signals to let drivers know where you're going, try to make eye contact with them and look before you go.

Explanation:

4 0

2 years ago

Read 2 more answers