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Simora [160]
4 years ago
15

A 36,287 kg truck has a momentum of 907,175 kg • . What is the truck’s velocity

Physics
2 answers:
yuradex [85]4 years ago
4 0

Momentum = (mass) x (speed) .

So 907,175 kg-m/s = (36,287 kg) x (speed)

Divide each side by (36,287 kg):

Speed = (907,175 kg-m/s) / (36,287 kg)

<em>Speed = 25 m/s</em>

That's the truck's <em>speed</em>. We can't do anything that involves the truck's <em>velocity,</em> because we don't know what direction the truck is moving.  

Serggg [28]4 years ago
4 0

The answer would be 25 m/s, I know because I just did it and got it right.

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Answer:

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Explanation:

8 0
3 years ago
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weight of Ali is 500andN.he is standing on the ground with an area of 0.025 m^2 area .we can find pressure under his feet. what
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Answer:

20000 Pa

Explanation:

Pressure is defined as the force per unit area.

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P= 500/0.025

P= 20000 Pa

8 0
3 years ago
If an object's mass increases, its--
aev [14]

Answer:

both kinetic and potential energy

Explanation:

this is your ans

I hope it helps mate

I will always help you understanding your assingments

have a great day

#Captainpower :)

8 0
3 years ago
The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2
Vlada [557]

Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x10^{-11} e^{-255/T}  and 2x10^{-12} e^{-940/T}  

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]  

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]

Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the  reaction with OH

Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x10^{-11} e^{-255/T}  and 2x10^{-12} e^{-940/T}  

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]  

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]

Than, considering the concentration of Cl increases by a factor of 100 to about 4 × 10^{5} molecules cm^{-3}, the result will be that the reaction with OH will be 535 + (100 to about 4 × 10^{5} molecules cm^{-3}) times faster than the  reaction with Cl

Explanation:

4 0
3 years ago
An ice cube at 0c was dropped into 30.0 g of water in a cup at 45.0c. at the instant that all of the ice was melted, the tempera
Ede4ka [16]
The amount of heat given by the water to the block of ice can be calculated by using
Q=m_w C_{sw} \Delta T_w
where 
m_w = 30 g is the mass of the water
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\Delta T_w = 45.0^{\circ}-19.5^{\circ}C = 20.5^{\circ}C is the variation of temperature of the water.

Using these numbers, we find
Q=(30 g)(4.18 J/(g^{\circ}C))(20.5^{\circ}C)=2571 J

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:
Q = m_i L_f
where m_i is the mass of the ice while L_f =334 J/g is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
m_i =  \frac{Q}{L_f}= \frac{2571 J}{334 J/g} =7.7 g
3 0
3 years ago
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