Answer:
9) This is a case of deceleration
10)-0.8 ms-2
b) acceleration is the change in velocity with time
11)
a) 100 ms-1
b) 100 seconds
12) 10ms-1
13) more information is needed to answer the question
14) - 0.4 ms^-2
15) 0.8 ms^-2
Explanation:
The deceleration is;
v-u/t
v= final velocity
u= initial velocity
t= time taken
20-60/50 =- 40/50= -0.8 ms-2
11)
Since it starts from rest, u=0 hence
v= u + at
v= 10 ×10
v= 100 ms-1
b)
v= u + at but u=0
1000 = 10 t
t= 1000/10
t= 100 seconds
12) since the sprinter must have started from rest, u= 0
v= u + at
v= 5 × 2
v= 10ms-1
14)
v- u/t
10 - 20/ 25
10/25
=- 0.4 ms^-2
15)
a=v-u/t
From rest, u=0
8 - 0/10
a= 8/10
a= 0.8 ms^-2
A. The particles are packed more tightly in materials with more density which causes the vibrations to bounce of the partials more rapidly which makes them go faster
Photochemical smog is formed when primary air pollutants interact with sunlight.
Photochemical smog is the result of the reaction between pollutants like nitrogen oxides (NO), sunlight and volatile organic compound (VOC) in the atmosphere. The sources of NO are car exhaust, coal power plants, factory emissions, etc. This type of smog is also known by the name Los Angeles smog.
Air pollutants are the particles present dissolved in the air, which when inhaled by the organisms can cause serious health issues. These pollutants are :ozone, particulate matter, gaseous oxides, etc. These pollutants majorly affect the respiratory system of the humans.
Therefore, photochemical smog is a form of pollution created when vehicle exhaust interacts with sunlight.
To know more about photochemical smog, here: brainly.com/question/15728274
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Answer: The distance is 723.4km
Explanation:
The velocity of the transverse waves is 8.9km/s
The velocity of the longitudinal wave is 5.1 km/s
The transverse one reaches 68 seconds before the longitudinal.
if the distance is X, we know that:
X/(9.8km/s) = T1
X/(5.1km/s) = T2
T2 = T1 + 68s
Where T1 and T2 are the time that each wave needs to reach the sesmograph.
We replace the third equation into the second and get:
X/(9.8km/s) = T1
X/(5.1km/s) = T1 + 68s
Now, we can replace T1 from the first equation into the second one:
X/(5.1km/s) = X/(9.8km/s) + 68s
Now we can solve it for X and find the distance.
X/(5.1km/s) - X/(9.8km/s) = 68s
X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s
X = 68s/0.094s/km = 723.4 km
Answer:3,600 Newtons
Explanation:
The net force acting on the car is
3×10^3squared
Newtons.
Force is defined as the product of the mass of the body and its aaceleration,⇒F=ma
Substituting the above given values we get,F=(1500kg) (2.0m /s^2 squared)=3000 N=3×10^3 squared N.
N=newtons
