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2 years ago

A truck speed from rest to 120km/h in 13s. Find its acceleration. (convert to m/s)

Physics

1 answer:

ioda2 years ago

4 0

answer:

2.5 m/s²

Explanation:

120 km/h = 120 ÷ 3.6 = 100/3 ≈ 33 m/s

a = (v2 - v1)/∆t = (33m/s - 0)/ 13s = 33/13 m/s²≈ 2.5 m/s²

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Nan investigated how many students at her high school had summer jobs. Which of the following is the procedure of her investigat

aalyn [17]

<span>In order to answer her question regarding how many students at her high school had summer jobs Nan must follow the procedure given in answer option B: First, randomly select 100 students, then ask them if they have summer jobs. In statistics this method is known as Population Sampling. These 100 students are chosen at random in order to avoid bias - we need them to be representative of the entire high school population.</span>

7 0

3 years ago

A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the ot

sp2606 [1]

Answer:

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

Initial momentum = Final momentum

4.4v = 12.364 i + 15.092 j

v =2.81 i + 3.43 j

Magnitude

$v=\sqrt{2.81^2+3.43^2}=4.43m/s$

Direction

$\theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0$

50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

6 0

3 years ago

2. A cinder block is sitting on a platform 20 m high. It weighs 16kg. The block has

Cerrena [4.2K]

Answer:

3136 Joules

Explanation:

Applying,

P.E = mgh.............. Equation 1

Where P.E = potential energy, m = mass of the cinder block, h = height of the platform, g = acceleration due to gravity.

From the question,

Given: m = 16 kg, h = 20 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

P.E = 16(20)(9.8)

P.E = 3136 Joules

Hence the potential energy of the cinder block is 3136 Joules

7 0

2 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 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]$

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 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]$

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

2 years ago

What is newtons 3rd law? pls help!

drek231 [11]

For every actions, there is an opposite reaction.

3 0

2 years ago

Read 2 more answers