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Mnenie [13.5K]
3 years ago
15

What is the torque τb about axis b due to the force f⃗ ? (b is the point at cartesian coordinates (0,b), located a distance b fr

om the origin along the y axis.) express the torque about axis b in terms of f, θ, ϕ, π, and/or other given coordinate data?

Physics
1 answer:
yKpoI14uk [10]3 years ago
8 0
Check the attached file for the solution.

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A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend
Viefleur [7K]

a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how

long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175

s this is the time to fall from the top; it would take the same time to travel

upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175

= 0.35s

b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice

to solve this problem the time it takes to fall the final 0.13 m is: time it

takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to

fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it

takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m

is then twice this, or 0.08s

5 0
4 years ago
Read 2 more answers
A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wave
Ann [662]

Answer:

\lambda = 40 cm

Explanation:

given data

string length = 30 cm

solution

we take here equation of length that is

L = n \times \frac{1}{4} \lambda     ...............1

so

total length will be here

L = \frac{\lambda}{2} +  \frac{\lambda}{4}\\

L = \frac{3 \lambda }{4}

so \lambda  will be

\lambda = \frac{4L}{3}\\\lambda = \frac{4\times 30}{3}

\lambda = 40 cm

5 0
3 years ago
Sarah walks WEST a distance of 5 meters then turns NORTH and walks 8
Leokris [45]

Answer:

she is 3 meters in north

Explanation:

3 0
3 years ago
A very humble bumble bee is flying horizontally due North at a constant speed of 3.11 m/s. At the current location of the bumble
Reil [10]

To solve this problem we will apply the concepts of the Magnetic Force. This expression will be expressed in both the vector and the scalar ways. Through this second we can directly use the presented values and replace them to obtain the value of the magnitude. Mathematically this can be described as,

\vec{F_B} = q(\vec{V}\times \vec{B})

F_B = q|v||B| sin\theta

Here,

q = Charge

v = Velocity

B = Magnetic field

\theta = \text{Angle between } \vec{B} \text{ and } \vec{V}

Our values are given as,

\theta = 35.7\°

q = 22.5*10^{-9}C

B = 1.05*10^{-5}T

v = 3.11m/s

Replacing,

F_B = (22.5*10^{-9}C)(3.11 \times 1.05*10^{-5}) sin(35.1\°)

F_B = 4.224*10^{-13}N

Therefore the size of the magnetic force acting on the bumble bee is 4.22*10^{-13}N

3 0
4 years ago
A rock thrown with speed 12.0 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 15.5 m bef
Dimas [21]

Supposing there's no air resistance, horizontal velocity is constant, which makes it very easy to solve for the amount of time that the rock was in the air.


Initial horizontal velocity is: <span>
cos(30 degrees) * 12m/s = 10.3923m/s 

15.5m / 10.3923m/s = 1.49s 

So the rock was in the air for 1.49 seconds. </span>

<span>

Now that we know that, we can use the following kinematics equation: 

d = v i * t + 1/2 * a * t^2 

Where d is the difference in y position, t is the time that the rock was in the air, and a is the vertical acceleration: -9.80m/s^2. </span>

<span>
Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s 

So: 

d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2 
d = 8.94 + -10.89</span>

d = -1.95<span>

<span>This means that the initial y position is 1.95 m higher than where the rock lands. </span></span>

5 0
3 years ago
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