I need these for test corrections that are due tomorrow please :)

Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin

likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0 and T - mgcos57 - F = m(0) = 0

Mg - T = 0 (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0

3 years ago

Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2

Korvikt [17]

Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

$x=v\cos\theta t$

On substituting the known values,

$\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}$

The change in the height, y of the basketball is given by,

$y=-v\sin\theta t+\frac{1}{2}gt^2$

Where g is the acceleration due to gravity.

On substituting the known values,

$\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}$

Final answer:

The parametric equations describing the shot are

$\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}$

8 0

1 year ago

Which of the following is a result of a change in pressure?

ioda

B. Exfoliation. Hope I helped you out bro.

7 0

3 years ago

If an incident light ray hits a flat smooth object at 28 degrees, it will reflect off at an angle of…

Nookie1986 [14]

Option B is correct. If an incident light ray hits a flat, smooth object at 28°. It will reflect off at an angle of 28°.

<h3>What is the law of reflection?</h3>

The law of reflection specifies that upon reflection from a downy surface, the slope of the reflected ray is similar to the slope of the incident ray.

The reflected ray is consistently in the plane determined by the incident ray and perpendicular to the surface at the point of reference of the incident ray.

When the light rays descend on the smooth surface, the angle of reflection is similar to the angle of incidence, also the incident ray, the reflected ray, and the normal to the surface all lie in a similar plane.

If an incident light ray hits a flat, smooth object at 28 degrees, it will reflect off at an angle of 28°.

Hence, option B is correct

To learn more about the law of reflection, refer to the link;

brainly.com/question/12029226

#SPJ1

5 0

1 year ago

A 1.00 kg object moving in the + x direction at 10.0 m/s collides with a 1.50 kg object traveling at 5.00 m/s in the - x directi

marysya [2.9K]

Answer:

The amount of kinetic energy lost during the collision is 60.75 J

Explanation:

The given parameters are;

The mass of the object moving in the +x direction, m₁ = 1.00 kg

The velocity of the object moving in the +x direction, v₁ = 10.0 m/s

The mass of the object moving in the -x direction, m₂ = 1.50 kg

The velocity of the object moving in the +x direction, v₂ = 5.00 m/s

The final velocity of the 1.00 kg mass after collision, v₃ = 4.00 m/s

The direction of motion of the 1.00 kg mass after collision = -x direction

The total initial kinetic energy of the system, $K.E._{total \ initial}$ = 1/2·m₁·v₁² + 1/2·m₂·v₂²

∴ $K.E._{total \ initial}$ = 1/2×1.00×10² + 1/2×1.50×5² = 68.75 J

The final kinetic energy of the system, $K.E._{final}$ = 1/2·m₁·v₃²

$K.E._{final}$ = 1/2×1.00×4² = 8 J

The amount of kinetic energy lost during the collision, $\Delta K.E._{system}$ is given as follows;

$\Delta K.E._{system}$ = $K.E._{final}$ - $K.E._{total \ initial}$ = 8 J - 68.75 J = -60.75 J

The amount of kinetic energy lost during the collision = 60.75 J.

7 0

2 years ago