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3 years ago

A rope horizontally pulls a massive object lying on a surface with friction with a constant

Physics

1 answer:

SVEN [57.7K]3 years ago

5 0

Answer:

Equal to the frictional force

Explanation:

Question; The options given with regards to a similar question posted online are;

A. Equal (equivalent) to the frictional force

B. Larger than the frictional force

C. Equal to the object's weight

D. More than the object's weight

Explanation

According to Newton's first law of motion, every object shall remain at rest or continue moving with uniform (constant speed) motion unless there is a net force acting on the object

Given that the velocity of the massive block, lying on the surface that has friction, being pulled by the rope = Constant

Therefore;

The net force acting on the moving block while being pulled by the rope = 0

From which we have;

The pulling force = The resistive force

Where;

The pulling force = The (pulling) force (applied) on the rope

The resistive force = The frictional force of the surface which tends to prevent the motion of the block

Therefore, given that the net force acting on the block = 0

The force on the rope = The frictional force (of the surface)

The correct option is option A. Equal to the frictional force.

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A crude approximation for the x component of velocity in an incompressible laminar boundary layer is a linear variation from u =

slega [8]

Answer:

2.5 * 10^-3

Explanation:

solution:

The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:

δu/δx+δv/δy=0

so that:

δv/δy= -δu/δx

Now, since u = Uy/δ, where δ = cx^1/2, we have that:

u=U*y/cx^1/2

and we obtain:

δv/δy=U*y/2cx^3/2

The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):

v=∫δv/δy(dy)=U*y/4cx^1/2

=y/x*(U*y/4cx^1/2)

=u*y/4x

which is exactly what we needed to demonstrate.

Also, using u = U*y/δ in the last equation we can obtain:

v/U=u*y/4*U*x

=y^2/4*δ*x

which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:

(v/U)_max=δ^2/4δx

=δ/4x

=2.5 * 10^-3

7 0

3 years ago

A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the

olasank [31]

Explanation:

The given data is as follows.

Spring constant (k) = 78 N/m, $\theta = 30^{o}$

Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

mgh sin $\theta = \frac{1}{2}kx^{2}$

h = $\frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}$

= $\frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}$

= 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.

4 0

3 years ago

Each plate of an air-filled parallel-plate capacitor has an area of 45.0 cm2, and the separation of the plates is 0.080 mm. A ba

maw [93]

Answer:

Option (e)

Explanation:

A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,

Energy density = 100 J/m

Let Q be the charge on the plates.

Energy density = 1/2 x ε0 x E^2

100 = 0.5 x 8.854 x 10^-12 x E^2

E = 4.75 x 10^6 V/m

V = E x d

V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V

C = ε0 A / d

C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F

Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C

Q = 190 nC

3 0

3 years ago

A woman with a mass of 52.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.470 rev/s abo

Charra [1.4K]

Explanation:

It is given that,

Mass of the woman, m₁ = 52 kg

Angular velocity, $\omega=0.47\ rev/s=2.95\ rad/s$

Mass of disk, m₂ = 118 kg

Radius of the disk, r = 3.9 m

The moment of inertia of woman which is standing at the rim of a large disk is :

$I={m_1r^2}$

$I={52\times 3.9^2}$

I₁ = 790.92 kg-m²

The moment of inertia of of the disk about an axis through its center is given by :

$I_2=\dfrac{m_2r^2}{2}$

$I_2=\dfrac{118\times (3.9)^2}{2}$

I₂ =897.39 kg-m²

Total moment of inertia of the system is given by :

$I=I_1+I_2$

$I=790.92+897.39$

I = 1688.31 kg-m²

The angular momentum of the system is :

$L=I\times \omega$

$L=1688.31 \times 2.95$

$L=4980.5\ kg-m^2/s$

So, the total angular momentum of the system is 4980.5 kg-m²/s. Hence, this is the required solution.

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3 years ago

According to the theory of plate tectonics, which forces cause the movement of plates in the earth's crust?

Sauron [17]

The correct answer is B.

4 0

3 years ago

Read 2 more answers