that is because in the gas particles there are more space while in the other hand why u can't compress liquid because most of the volume of a gas is composed of the large amounts of empty space between the gas particles.
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
Answer:
The coefficient of kinetic friction between the block and the surface is 0.127.
Explanation:
Given that,
The mass of a block, m = 4 kg
The acceleration of the block, a = -1.25 m/s²
We need to find the coefficient of kinetic friction between the block and the surface. The force of friction is given by :
So, the coefficient of kinetic friction between the block and the surface is 0.127.
Answer:
P = 25299.75 watts
Since 80km/h is the average speed of 92km/h and 68km/h, the power (in watts) is needed to keep the car traveling at a constant 80 km/h is P = 25299.75 watts
Explanation:
Given;
Mass of car m = 1280kg
initial speed v1 = 92km/h = 92×1000/3600 m/s= 25.56m/s
Final speed v2 = 68km/h = 68×1000/3600 m/s= 18.89m/s
time taken t = 7.5s
Change in the kinetic energy of the car within that period;
∆K.E = 1/2 ×mv1^2 - 1/2 × mv2^2
∆K.E = 0.5m(v1^2 -v2^2)
Substituting the values, we have;
∆K.E = 0.5×1280(25.56^2 - 18.89^2)
∆K.E = 189748.16J
Power used during this Change;
Power P = ∆K.E/t
Substituting the values;
P = 189748.16/7.5
P = 25299.75 watts
Since 80km/h is the average speed of 92km/h and 68km/h, the power (in watts) is needed to keep the car traveling at a constant 80 km/h is P = 25299.75 watts
Answer:
The horizontal component of the velocity vector is;
vh = 34.4 ft/s
The vertical component of the velocity vector is;
vy = 49.1 ft/s
Explanation:
Given;
Velocity of football v = 60 ft/s
Angle of elevation ∅ = 55°
The horizontal component of the velocity vector is;
vh = vcos∅
Substituting the values;
vh = 60cos55°
vh = 34.41458618106 ft/s
vh = 34.4 ft/s
The vertical component of the velocity vector is;
vy = vsin∅
Substituting the values;
vy = 60sin55°
vy = 49.14912265733 ft/s
vy = 49.1 ft/s
