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2 years ago

14

A 2.0-kilogram cantaloupe rolling north at 4.0 meters per second collides head on with a 1.0-kilogram orange rolling south at 8.

0 meters per second. What is the magnitude of the total momentum of the two fruit?

2 answers:

Murrr4er [49]2 years ago

8 0

Answer:

Hi how are you doing today Jasmine

Nat2105 [25]2 years ago

7 0

Answer:

16kg*m/s

Explanation:

a)p=m*v=2*4=8kg-m/s

b)p=m*v=1*8=8kg-m/s

add them

=16kg-m/s

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a 1.50*10^-5 C charge feels a 2.89*10^-3 N force when it moves 288m/s perpendicular (90) deg to a magnetic field. how strong is

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6.68, -1
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3 years ago

A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subj

KATRIN_1 [288]

Answer:

$\Delta t = 8 s$

Explanation:

As we know that the angular acceleration of the wheel due to friction is constant

so we can use kinematics

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

so we have

$(65 \times 2\pi) = (2\pi \times 9)(10) + \frac{1}{2}(\alpha)(10^2)$

$130\pi = 180\pi + 50 \alpha$

$\alpha = -\pi rad/s^2$

now time required to completely stop the wheel is given as

$\omega_f = \omega_i + \alpha t$

$0 = (2\pi \times 9) + (-\pi) t$

$t = 18 s$

now time required to stop the wheel is given as

$\Delta t = 18 - 10$

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6 0

3 years ago

Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. Take T = 1.3 kN·m. De

My name is Ann [436]

Answer:

τ (bc. max) =25.37 MPa

Explanation:

From the question, T = 1.3Kn.m;

(G = 77.2 GPa) and from the image of this solid shaft system i attached;

d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m

So ΣT = 0 → Ta + Tc = 1.3Kn.m

So the system is statically indeterminate.

Let's check at the equation that makes it compatible ;

ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0

[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =

ΣT = 0 and T(bc) = T(c)

ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m

Now,

[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0

So, T(c) = 273374 Nmm = 273.374Nm

T(a) = 1300Nm - 273.374Nm = 1026. 63Nm

From the beginning we saw that;

T(bc) = T(c) = 273.374Nm

Now let's find the maximum shear stress in shaft BC;

τ (bc. max) = {τ (bc) x r(bc)} / J(bc)

τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707

= 25.37 MPa

3 0

3 years ago