A 500 gram mass attached to a horizontal spring

HELP PLEASE!! An illustration of a pendulum at 3 positions of a pendulum. The equilibrium position is labeled B. The other two p

Sedbober [7]

Answer:

Both A and C

Explanation:

I just got it correct on Edg

7 0

2 years ago

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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

8 months ago

Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a

lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0

3 years ago

What are the opposite ends of a magnet called? A. north and south terminals B. north and south poles C. magnetic fields D. magne

Bezzdna [24]

I think the answer is B.

Hope this helps.

6 0

2 years ago

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Ahmad is riding his bicycle. He finds that he can accelerate from rest at 0.44 m/s^2 for 5 s to reach a speed of 2.2 m/s. The to

snow_lady [41]

Answer:

1) The force Christian can exert on his bicycle before picking up the the cargo is 529.74 N

2) The force Christian can exert on his bicycle after picking up the the cargo is 647.46 N

Therefore, Christian has to exert more force on his bike after picking up the cargo

Explanation:

The given parameters are;

The mass of Christian and his bicycle = 54 kg

The mass of the cargo = 12 kg

1) The force Christian can exert on his bicycle before picking up the the cargo = Mass of Christian and his bicycle × Acceleration due to gravity

∴ The force Christian can exert on his bicycle before picking up the the cargo = 54 kg × 9.81 m/s² = 529.74 N

2) The force Christian can exert on his bicycle after picking up the the cargo = (54 + 12) kg × 9.81 m/s² = 647.46 N

Therefore, Christian has to exert more force on his bike after picking up the cargo.

7 0

2 years ago