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lesya [120]
4 years ago
11

driving down the highway , you find yourself behind a heavily loaded tomato truck. you follow close behind the truck keeping the

same speed. suddenly a tomato falls from the back of the struck. will the tomato hit your car or land on the road assuming you continue moving the same speed and direction.
Physics
1 answer:
prohojiy [21]4 years ago
4 0

Answer:

The tomato won't hit the car

Explanation:

According to the statement, the car moves at constant speed behind the truck fully loaded with tomatoes, and in the same direction. When a tomato falls from the top of the truck, it should not hit the car as the tomato falls due to the force of gravity, while horizontally has the same speed and in the same direction as the truck.  So we assume that the tomato will fall to the road without touching the car.

Have a nice day!

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The acceleration of the electron is calculated as follows;

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a = (1.6 x 10⁻¹⁹ x 320)/(9.11 x 10⁻³¹)

a = 5.62 x 10¹³ m/s²

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v = at

v = 5.62 x 10¹³ m/s² x  8.50 x 10⁻⁹ s

v = 4.78 x 10⁵ m/s

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2 years ago
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An object is stopped at point ‘a’, and travels to point ‘c’ in 8s. What was its acceleration?
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7 0
3 years ago
A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate
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Answer: The molar heat capacity of aluminum is 25.3J/mol^0C

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of water = 130.0 g

m_2 = mass of aluminiunm = 23.5 g

T_{final} = final temperature = 26.0^oC=(273+26)K=299K

T_1 = temperature of water = 23^oC=(273+23)K=296K

T_2 = temperature of aluminium = 100^oC=273+100=373K

c_1 = specific heat of water= 4.184J/g^0C

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Now put all the given values in equation (1), we get

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5 0
3 years ago
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Answer:

0.005734 s^{-1} and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

Explanation:

Expression for rate law for first order kinetics is given by:

a_o=a\times e^{-kt}

where,

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We have :

a_o=x

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t = 95 s

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Half life is given by for first order kinetics::

t_{1/2}=\frac{0.693}{k}

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0.005734 s^{-1} and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

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3 years ago
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