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hoa [83]
3 years ago
6

Flash floods can cause vehicles to float and fill with water, trapping and drowning people. while especially dangerous at night

and in deep water, even ____ inches of water can float some small cars.
a. two
b. four
c. six
d. eight
Physics
1 answer:
IceJOKER [234]3 years ago
3 0
I think the answer is: 6 inches

On average, the distance between Car's engine to the ground is around 8-10 inches.. But for small cars, the distance between car's engine to the ground could be as low as 5 inches - 6 inches from the ground.

When the flood reach the car's engine level, the water will got into the engine and disrupt it function, causing the car became unable to work properly
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A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi
bogdanovich [222]

Answer:

1) W_{object} = 400 N

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

\sum F_{Tot} = 0

In this case we just have y-direction forces.

\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0

Now, let's solve the equation for W(object).

W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N

2 ) To find the position of the heaviest weight we need to use the torque definition.

\sum \tau = 0

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

L = 1.50 m  ; total length of the bar.

D_{AP} = x  ; distance between Tension A and P point.

D_{BP} = L-x ; distance between Tension B and P point.

D_{W_{bar}P} = \frac{L}{2}-x ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0

Finally we just need to solve it for x.

x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}

x = 0.28 m

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0
4 years ago
A coil 3.55 cm in radius, containing 470 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.2
slavikrds [6]

The Electric current is 1.11* 10^{-4}A


Given that the coil's radius is 3.55 cm (0.35 m),

The formula for the coil's area is A = r2 A = (3.14) (0.35)2 = 0.005024 m2.

R = Resistance = 600 N = Number of spins = 500 B = Magnetic field = (0.0120)

t + (3 x 10⁻⁵) t⁴

The number t = 5 is substituted for taking the derivative at both the induced current and the electric current.

The Electric current is therefore 1.11* 10^{-4}A
Electric current - The rate of electron passage in a conductor is known as electric current. The ampere is the electric current's SI unit. Electrons are little particles that are part of a substance's molecular structure. These electrons can be held loosely or securely depending on the situation.

To learn more about electric current please visit -brainly.com/question/12791045
#SPJ1

7 0
2 years ago
Read 2 more answers
What cloud makes hail
MArishka [77]

Answer:

Cumulonimbus

Hail development. Hail is a type of strong precipitation, which is shaped in rainstorms mists (Cumulonimbus). Tempests mists comprises of beads of fluid water (at temperatures lower than 0°с, the beads can be in a thermodynamically instable supercooled condition) and ice gems

Explanation:

3 0
3 years ago
You want to find out how many atoms of the isotope 65Cu are in a small sample of material. You bombard the sample with neutrons
serious [3.7K]

Answer:

a) number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶ atoms

b) m_total Cu = 1.585 10⁹ u = 2.632 10⁻¹⁸ kg

Explanation:

a) For this exercise let's start by using the radioactive decay ratio

           N = N₀  e^{- \lambda t}o e - lambda t

The half-life time is defined as the time it takes for half of the radioactive (activated) atoms to decay, therefore after two half-lives there are

            N = ½ (½ N₀) = ¼ N₀

            N₀ = 4 N

in each decay a photon is emitted so we can use a direct rule of proportions. If an atom emits a photon it has Eo = 1,04 Mev, how many photons it has energy E = 10,000 MeV

          # _atoms = 1 atom (photon) (E / Eo)

          # _atoms = 1 10000 / 1.04

          # _atoms = 9615,4 atoms

          N₀ = 4 #_atoms

          N₀ = 4 9615,4

          N₀=  38461.6  atoms

in the exercise indicates that half of the atoms decay in this way and the other half decays directly to the base state of Zinc, so the total number of activated atoms

          N_activated = 2 # _atoms

          N_activated = 2 38461.6

          N_activated = 76923.2

also indicates that 1% = 0.01 of the nuclei is activated by neutron bombardment

          N_activated = 0.01 N_total

          N_total = N_activated / 0.01

          N_total = 76923.2 / 100

          N_total = 7.692 10⁶ atoms

so the number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶

b) the natural abundance of copper is

  ⁶³Cu     69.17%

  ⁶⁵Cu    30.83%

Let's use a direct proportion rule. If there are 7.692 10⁶  ⁶⁵Cu that represents 30.83, how much ⁶³Cu is there that represents 69.17%

                # _63Cu = 69.17%  (7.692 10⁶    / 30.83%)

                # _63Cu = 17.258 10⁶  atom  ⁶³Cu

the total amount of comatose is

              #_total Cu = #_ 65Cu + # _63Cu

              #_total Cu = (7.692 + 17.258) 10⁶

              #_total Cu = 24.95 10⁶

the atomic mass of copper is m_Cu = 63.546 u

          m_total = #_totalCu m_Cu

          m_total = 24.95 10⁶ 63,546 u

          m_total = 1.585 10⁹ u

let's reduce to kg

           m_total Cu = 1.585 10⁹ u (1,66054 10⁻²⁷ kg / 1 u)

           m_total Cu = 2.632 10⁻¹⁸ kg

8 0
3 years ago
A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a cons
torisob [31]

Answer:

0.1 s

Explanation:

The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log

So F - f = ma

F - μmg = ma

F/m - μg = a

So, substituting the values of the variables into the equation, we have

a = F/m - μg

a = 2850 N/300 kg - 0.45 × 9.8 m/s²

a = 9.5 m/s² - 4.41 m/s²

a = 5.09 m/s²

Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.

So, making t subject of the formula, we have

t = (v - u)/a

substituting the values of the variables into the equation, we have

t = (v - u)/a

t = (0.5 m/s - 0 m/s)/5.09 m/s²

t = 0.5 m/s ÷ 5.09 m/s²

t = 0.098 s

t ≅ 0.1 s

6 0
3 years ago
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