For mach index of 0.6 what is the intake valve flow area g

a student moves a box across the floor by exerting 23.3 N of force and doing 47.2 J of work on the box. How far does the student

kozerog [31]

Work = Force x Distance
47.2J = 23.3N x d
d = 47.2/23.3
d = 2.0258 m

hope this helps :P

8 0

3 years ago

What is the law of gravity?

Harlamova29_29 [7]

Newton's law of universal gravitation states that a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them

7 0

3 years ago

5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley

ella [17]

Answer:

1) Please find attached, created with Microsoft Visio

2) The acceleration of the masses connected by the light string is 0.00735 m/s²

3) The tension in the cord is 0.147 N

4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s

5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s

Explanation:

1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio

2) The acceleration of the masses connected by the light string is given as follows;

F = Mass, m × Acceleration, a

The mass of the truck, M = 20.0 kg

The mass attached to the string, hanging rom the pulley, m = 0.0150 kg

The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N

The pulling force acting on the cart, F = M × a

∴ F = 0.147 N = 20.0 kg × a

a = 0.147 N/(20.0 kg) = 0.00735 m/s²

The acceleration of the truck = a = 0.00735 m/s²

3) The tension in the cord = F = 0.147 N

4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²

Where;

s = The distance to the edge of the table = 1.2 m

u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)

a = The acceleration of the cart = 0.00735 m/s²

t = The time taken

Substituting the known values, gives;

s = u·t + 1/2·a·t²

1.2 = 0 × t + 1/2 ×0.00735 × t²

1.2 = 1/2 ×0.00735 × t²

t² = 1.2/(1/2 ×0.00735) ≈ 326.5306

t = √(1.2/(1/2 ×0.00735)) ≈ 18.07

The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s

5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;

v² = u² + 2·a·s

u = 0 m/s

v² = 0² + 2 × 0.00735 × 1.2 = 0.001764

v = √(0.001764) = 0.042

The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.

7 0

3 years ago

Read 2 more answers

What is the wavelength of light that is deviated in the first order through an angle of 13.1 ∘ by a transmission grating having

faltersainse [42]

Answer:

The wavelength of light is $4.53\times10^{-7}\ m$

Explanation:

Given that,

Angle = 13.1°

Number of slits = 5000

We need to calculate the wavelength of light

Diffraction of first order is defined as,

$d \sin\theta=n\lambda$ .....(I)

The separation of the slits

$d = \dfrac{1}{N}$

$d=\dfrac{1}{5000}$

$d=2\times10^{-6}\ m$

Now put the value in equation (I)

$2\times10^{-6}\sin13.1^{\circ}=\lambda$

Here, n = 1

$\lambda=4.53\times10^{-7}\ m$

Hence, The wavelength of light is $4.53\times10^{-7}\ m$

7 0

4 years ago

Part C: Quantitative Problems when vf is not 0

Alina [70]

Answer:

(a)

$\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s$

(b)

1120 N

Explanation:

Change in velocity, $\triangle v$ is given by subtracting the initial velocity from the final velocity and expressed as $\triangle v= v_f -v_i$

Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

$\triangle v=0-8=-8\ m/s$