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2 years ago

A 25-kilogram mass travels east with a constant velocity of 40. meters per second. the momentum of this mass is

2 answers:

4 0

<span>Momentum, a quantity of motion of a moving body, is mathematically defined as mass x velocity. Hence the magnitude of momentum of a 25-kg mass traveling at 40 m/s is 40x25 = 1000 kg-m/s. As momentum is directional in nature, momentum = (1000i + 0j + 0k) kg-m/s in vector notation (as body is traveling east)</span>

DiKsa [7]2 years ago

3 0

Momentum is the product of speed/velocity and the mass of a body in motion. The momentum of a body at rest is zero since its speed is also zero. In this case, the mass is 25 kg while the velocity is 40 m/s.
Therefore, momentum = mass x velocity
= 25 × 40
= 1000 kgm/s

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The 10-lb block has a speed of 4 ft/s when the force of f=(8t2)f=(8t2) lb is applied. determine the velocity of the block when t

KatRina [158]

The velocity of the block when t == 2 s is 60.7 ft./sec.

Equations of Motion.

Here the friction is $F_f = \mu_k N$ = 0.2 N

$+ \uparrow \sum F_y = ma_y; \quad N – 10 = \frac { 10 } { 32.2 }(0) \quad N = 10 lb \\ \begin{aligned} \underrightarrow{ + } \sum F_x = ma_x; \quad 8t^2 – 0.2(10 &) = \frac { 10 } { 32.2 }a \\ & a = 3.22(8t^2 – 2) ft/s^2 \end{aligned}$

Kinematics.

The velocity of the block as a function of t can be determined by

integrating dv = adt using the initial condition v = 4 ft./s at t = 0.

$\int_{ 4 ft/s }^{ v } dv = \int_0^t 3.22(8t^2 – 2)dt \\ \begin{aligned} v – &4 = 3.22 (\frac 8 3 t^3 – 2t) \\ & v = \{8.5867t^3 – 6.44t + 4 \} ft/s \end{aligned}$

The displacement as a function of t can be determined by integrating

ds = vdt using

the initial condition s = 0 at t = 0

$\int_0^s ds = \int_0^t (8.5867t^3 – 6.44t + 4)dt \\ s = \{2.1467t^4 – 3.22t^2 + 4t \} ft$

at t = 2 sec

s = 30 ft.

Thus, at s = 30 ft.,

$\begin{aligned} v &= 8.5867(2.0089^3) – 6.44(2.0089) + 4 \\ &= 60.67 ft/s \\ &= 60.7 ft/s \end{aligned}$

Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that cause them to move.

Kinematics, as a field of study, is often referred to as the "geometry of motion" and is occasionally seen as a branch of mathematics. A kinematics problem begins by describing the geometry of the system and declaring the initial conditions of any known values of position, velocity and/or acceleration of points within the system.

Then, using arguments from geometry, the position, velocity and acceleration of any unknown parts of the system can be determined. The study of how forces act on bodies falls within kinetics, not kinematics. For further details, see analytical dynamics.

Learn more about kinematics here : brainly.com/question/24486060

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5 0

9 months ago

A rod 7.0 m long is pivoted at a point 2.0 m from the left end. A downward force of 50 N acts at the left end, and a downward fo

kicyunya [14]

If the rod is in rotational equilibrium, then the net torques acting on it is zero:

∑ τ = 0

Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:

• at the left end,

τ = + (50 N) (2.0 m) = 100 N•m

• at the right end,

τ = - (200 N) (5.0 m) = - 1000 N•m

• at a point a distance d to the right of the pivot point,

τ = + (300 N) d

Then

∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0

⇒ (300 N) d = 1100 N•m

⇒ d ≈ 3.7 m

6 0

2 years ago

Joe and Max shake hands and say goodbye. Joe walks east 0.40 km to a coffee shop, and Max flags a cab and rides north 3.65 km to

katen-ka-za [31]

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

$OB=0.40 km$

And the Max distance towards bookstore is,

$OA=3.65 km$

Now the distance between the Joy and Max will be,

By applying pythagorus theorem,

$AB=\sqrt{OB^{2}+OA^{2}}$

Substitute 0.40 km for OB and 3.65 km for OA in the above equation.

$AB=\sqrt{0.40^{2}+3.65^{2}}\\AB=\sqrt{13.4825} \\AB=3.67 km$

Therefore the distance between there destination is 3.67 km.

6 0

3 years ago

Gravity is affected by...

goldfiish [28.3K]

Mass and distance
force /pull of gravity decreases with the increase in separation between the two bodies
the amount of gravity an object possesses is proportional to the mass of that object.

4 0

3 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum spee dof 32.0 m/s. With what maximum sp

Dvinal [7]

Answer

Maximum speed at 75 m radius will be 22.625 m /sec

Explanation:

We have given radius of the curve r = 150 m

Maximum speed $v_{max}=32m/sec$