The force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.
<h3>What is force?</h3>
Force is defined as the product of mass and acceleration of an object.
Friction is defined as the force that resists the movement of an object over another.
Therefore, the force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.
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Answer:
<em>The correct option is D) fruit</em>
Explanation:
Both the gymnosperm and the angiosperms plants produce seed. Seeds are an essential part of reproduction in plants.
After fertilization has occurred, the ripened ovule is the part which turns into a seed. The size of the seeds depends on the type if plant.
After fertilization, the ovary forms the fruit. As the ovule is present in the ovule hence after fertilization, the seed which was made from the ovule gets enclosed in the fruit which was made from the ovary.
Answer:
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Explanation:
This is an angular kinematic exercise the equation for the angular position
the particle A
θ = θ₀ + ω₀ t + ½ α t²
They say for the particle B
w₀B = ½ w₀
αB = 2 α
In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial
t´ = t - t_1
l
et's write the equation of particle B
θ = θ₀ + w₀B t´ + ½ αB t´2
replace
θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Answer:
option C
Explanation:
Final velocity of the object is 114 m / s. Hence, final velocity of the object is 114 m / s.
Answer:
B. About 12 degrees
Explanation:
The orbital period is calculated using the following expression:
T = 2π*(
)
Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.
Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.
Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:
= 
Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be 
years, or 31.62 years.
We find the amount of degrees it moves in 1 year:
or about 12 degrees.
