Stars that are not very hot but give off a lot of light are
O nebula
O main sequence stars
giants
O giants
O dwarfs
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Giants.
thankshope it helps.Explanation:
giants are those stars that are not so hot but give a lot of light
use a trigonometric equation to determine the leg of this triangle
C=90°
A=30°
c=10m
What is a?
Answer: 5
Explanation: B is for sure 60°, c* cosB = 10*1/2 =5
Please help I’ll mark you brainliest
Answer:
Percentage:
Rr = 50% because it's 2/4 (for both or 25% each since you have them separate)
rr = also 50%, because it's also 2/4.
Phenotype:
Rr = heterozygous
rr = "hozygous" recessive
In addition, RR is "hozygous" dominant
Explanation:
They said the hozygous is a swearword LOL.
What is the centripetal force for a roller coaster if the mass is 10 kg and the normal force is 25 N?
Answer:
Fc = 123 Newton
Explanation:
Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.
Mathematically, net force is given by the formula;
[tex] Fnet = Fapp + Fg[/tex]
Where;
Fnet is the net force.
Fapp is the applied force.
Fg is the force due to gravitation.
Given the following data;
Normal force = 25N
Mass = 10kg
To find the centripetal force;
From the net force, we have the following formula;
Fc = N + mg
Where;
Fc is the centripetal force.
N is the normal force.
mg is the the weight of the object.
Substituting into the formula, we have;
Fc = 25 + 10(9.8)
Fc = 25 + 98
Fc = 123 Newton
A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of 18kg⋅m2. She then tucks into a small ball, decreasing this moment of inertia to 3.6kg⋅m2. While tucked, she makes two complete revolutions in 1.2s.
Required:
If she hadn't tucked at all, how many revolutions would she have made in the 1.5 s from board to water?
Answer:
θ₁ = 0.5 revolution
Explanation:
We will use the conservation of angular momentum as follows:
[tex]L_1=L_2\\I_1\omega_1=I_2\omega_2[/tex]
where,
I₁ = initial moment of inertia = 18 kg.m²
I₂ = Final moment of inertia = 3.6 kg.m²
ω₁ = initial angular velocity = ?
ω₂ = Final Angular velocity = [tex]\frac{\theta_2}{t_2} = \frac{2\ rev}{1.2\ s}[/tex] = 1.67 rev/s
Therefore,
[tex](18\ kg.m^2)\omega_1 = (3.6\ kg.m^2)(1.67\ rev/s)\\\\\omega_1 = \frac{(3.6\ kg.m^2)(1.67\ rev/s)}{(18\ kg.m^2)}\\\\\omega_1 = \frac{\theta_1}{t_1} = 0.333\ rev/s\\\\\theta_1 = (0.333\ rev/s)t_1[/tex]
where,
θ₁ = revolutions if she had not tucked at all = ?
t₁ = time = 1.5 s
Therefore,
[tex]\theta_1 = (0.333\ rev/s)(1.5\ s)\\[/tex]
θ₁ = 0.5 revolution