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Natasha_Volkova [10]
3 years ago
12

two individuals start the training program at the same time, but one is able to grow muscle faster an larger than other. what is

the least likely explanation for this?
Physics
1 answer:
Pavel [41]3 years ago
4 0

A.

One individual was born with more muscle fibers.

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Juan whose weight is 500 N is standing on the ground. The force the ground exerts on
Anna71 [15]

Answer:

more than 500 n i think the answer will

8 0
3 years ago
Read 2 more answers
uppose a wheel is initially rotating at 10.0 rad/s while undergoing constant angular acceleration reaching a speed of 30.0 rad/s
Paul [167]

Answer:

t = 22.2 s

Explanation:

angular distance covered in the 36.0 s is

θ = ω(avg)t = ½(10.0 + 30.0)36 = 720 radians

720/2 = 360 radians

α = Δω/t = (30 - 10)/36 = 5/9 rad/s²

θ = ω₀t + ½αt²

360 = 10.0t + ½(5/9)t²

   0 = (5/18)t² + 10.0t - 360

   0 = t² + 36t - 1296

t = (-36 ±√(36² - 4(1)(-1296))) / 2

t = (-36 ±√(6480)) / 2

t = -18 ±√1620

we ignore the negative time result as it occurs before we care.

t = -18 + √1620 = 22.249223... s

8 0
3 years ago
During a 0.001 s interval while it is between the plates, the change of the momentum of the electron Δp with arrow is < 0, -8
Delvig [45]

Answer:

5.5 x 10^5 N/C

Explanation:

t = 0.001 s

Δp = - 8.8 x 10^-17 kg m /s

Force is equal to the rate of change of momentum.

F = Δp / Δt

F = (8.8 x 10^-17) / 0.001 = 8.8 x 10^-14 N

q = 1.6 x 10^-19 C

Electric field, E = F / q = (8.8 x 10^-14) / (1.6 x 10^-19)

E = 5.5 x 10^5 N/C

8 0
4 years ago
Two particles, each with charge Q, and a third charge q, are placed at the vertices of an equilateral triangle as shown. The tot
Gelneren [198K]

Answer:

<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>

Explanation:

The image is shown below.

The force on the particle with charge q due to each charge Q = \frac{kQq}{r^{2} }

we designate this force as N

Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.

Resolving the forces on the particle, we have

for the x-component

N_{x} = N cosine 60° + (-N cosine 60°) = 0

for the y-component

N_{y} = -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N

The above indicates that there is no resultant force in the x-axis, since it is equal to zero (N_{x} = 0).

The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.

<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>

5 0
4 years ago
Are the two expressions equvalent 7(8x+5) and 48x + 35
Evgesh-ka [11]
Yes they are equivalent because 7x5=35 and 8x x 5=48x
4 0
4 years ago
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