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Ede4ka [16]
1 year ago
14

Assuming the pick-up trucks, trailers and road conditions are exactly the same, which vehicle will take a longer distance to sto

p?
Physics
1 answer:
mars1129 [50]1 year ago
6 0

I’ve answered this before so I know the question is missing an important given and that given is: <span>1 has an empty trailer and the other has a fully loaded one. 

So, it would be the fully loaded trailer that would take a longer distance to stop because a lot of weight is being pulled, and when the brakes are started, the fully loaded trailer is more like pushing against the truck.</span>

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The gravitational field of m1 is denoted by g1. Enter an expression for the gravitational field g1 at position la in terms of m1
Andrei [34K]

Answer:

The expression of gravitational field due to mass m_! at a distance l_a

Explanation:

We have given mass is m_1

Distance of the point where we have to find the gravitational field is l_a

Gravitational constant G

We have to find the gravitational filed

Gravitational field is given by g=\frac{Gm_1}{l_a^2}

This will be the expression of gravitational field due to mass m_! at a distance l_a

4 0
2 years ago
A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along
storchak [24]

Answer:

v = \frac{kQ}{a}  

Explanation:

We define the linear density of charge as:

\lambda = \frac{Q}{L}

     Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

dv = \frac{kdq}{r}

          where k is the coulomb's constant

                     r is the distance from dq to center of the circle

Thus.

v = \int_{}^{}\frac{kdq}{a}  

v = \frac{k}{a}\int_{}^{}dq

v = \frac{kQ}{a}     Potential at the center of the semicircle

4 0
1 year ago
In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.05 kg. The masses of the pulley and string are negligible by comparis
Rus_ich [418]
M1 descending
−m1g + T = m1a 

m2 ascending
m2g − T = m2a

this gives :
(m2 − m1)g = (m1 + m2)a 

a = (m2 − m1)g/m1 + m2
   = (5.60 − 2)/(2 + 5.60) x 9.81 
   = = 4.65m/s^2
5 0
1 year ago
A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glo
padilas [110]

Answer:

 a)    λ = 189.43 10⁻⁹ m  b)    λ = 269.19 10⁻⁹ m

Explanation:

The diffraction network is described by the expression

      d sin θ= m λ

Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

        tan θ = y / L

The diffraction spectrum is measured at very small angles, therefore

      tan θ = sin θ / cos θ = sin θ

We replace

      d y / L = m λ

Let's place in the first order m = 1

Let's look for the separation of the lines (d)

     d = λ  L / y

     d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²

     d = 332.33 10⁻⁹ m

Now we can look for the wavelength of the other line

     λ  = d y / L

    λ  = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²

    λ = 189.43 10⁻⁹ m

Part B

The compound wavelength B

      λ  = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

      λ = 269.19 10⁻⁹ m

4 0
2 years ago
the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point
Andreyy89

Answer:

The pressure at this point is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

P=1.81\ kW

We need to calculate the pressure at this point

Using formula of pressure

P=F\times v

P=A_{1}P_{1}\times v_{1}

Here, A_{1}v_{1}=Q_{1}

Where, v = velocity

Put the value into the formula

1.81 =P_{1}\times0.124\times\dfrac{1}{60}

P_{1}=\dfrac{1.81\times60}{0.124}

P_{1}=875.80\ kPa

P_{1}=0.875\ mPa

Hence, The pressure at this point is 0.875 mPa

5 0
1 year ago
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