An ant will have higher momentum than elephant only if elephant is standing still.
Explanation:
The word momentum is defined as product of object’s mass and velocity
buy which it moves and the momentum can be denoted by vector quantity
.
Whose product of mass and the velocity is more; its momentum will be more.
If elephant stands still, means its velocity is equal to zero, then the product of its mass and velocity will be zero. Hence, it will have 0 momentum.
Since, the mass of ant is insignificantly small but have velocity for moving. So, the ant have small momentum but somewhat greater than zero.
Therefore, the ant will have higher momentum than elephant only if elephant stands still.
If an ant will stand still it have zero velocity, so the momentum will be equal to zero. Hence, this option is also incorrect.
The option “The elephant has much high mass” is incorrect because having high mass but zero velocity will result in no momentum.
Thus, an ant will have higher momentum than elephant only if elephant is standing still.
Learn More:
1. Momentum <u>brainly.com/question/11942991 </u>
2. More about momentum <u>brainly.com/question/2872275 </u>
3. Force and motion <u>brainly.com/question/11947870 </u>
Answer Details:
Grade: High School
Subject: Physics
Chapter: Momentum
Keywords:
Ant, elephant, momentum, more, explain, standing, mass, fast, still, change, object, velocity, force, times, impulse, acting, negligible, product, zero.
Answer:
Mass of astronaut will be 119.7674 kg
Explanation:
We have given mass of the chair
When there is chair alone then time period of osculation
Now when astronaut sit on the chair then total mass
Now with astronaut time period is
For spring time period is given by
So
Squaring both side
Mass of astronaut will be 119.7674 kg
Answer:
=
Explanation:
given data:
Pressure P = 3 * 10^ 5 Pa
speed v = 5 m / s
Area A = A
from the information given in equation final Area is 1/3 of initial area i.e.
A ' = A / 3
we know that density of water = 1000 kg / m^ 3
from continuity equation
Av = A ' v'
so we have
speed v ' = 3*A*v / A
v ' = 3*A*5/ A
v = 15 m / s
from bernoulli's equation we can calculate final pressure
Required pressure P ' = P + ( 1/ 2) \rho [ v^ {2} v'^{ 2}]
=
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Answer:
The percent of power extended against sea-level air drag is 50.15%
Explanation:
Length L is 22 ft, Power P is 590 hp,
Speed of air V = 175 mi/h. To convert to ft, we will multiply by 1.46
V = 175 × 1.46 = 255.5 ft
Frontal Area A is given as 23.5 ft
Coefficient of drag, Cd is 0.35
From text, Density of Air at sea level ρ = 0.00237 slug/ft³
The formula for calculating Drag Force F is
F = Cd × × ρAV²
Substitute the values above into the formula
F = 0.35 × × 0.00237 × 23.5 × 255.5²
= 0.35 × × 3635.78
= 636.26 lbf
But Power P = FV
Therefore we substitute for F and V to get P
P = 636.26 × 255.5
= 162564.43 ft.lbf/s
To convert to horsepower, we multiply by 0.00182 hp
= 162564.43 × 0.00182
= 295.87 hp
The percent of power extended against sea-level air drag is
= (295.87/590) × 100
= 50.15%