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3 years ago

8

A 250-lb block is subjected to a horizontal force P. The coefficient of friction between the block and surface is µs = 0.2. Dete

rmine the force P required to start moving the block up the incline

1 answer:

aev [14]3 years ago

6 0

Answer:

force required to push the block = 219.714 lb

Explanation:

GIVEN DATA:

weight W of block = 250 lb

coefficient of friction = 0.2

consider equilibrium condition in x direction

$P*cos(30)-W*sin(30)-\mu _{s}N = 0$

$P*0.866-0.2N = 125$ .........................(1)

consider equilibrium condition in Y direction

$N-Wcos(30)-Psin(30)= 0$

$N-0.5P=216.503$ .....................(2)

SOLVING 1 and 2 equation we get N value

N = 326.36 lb

putting N value in either equation we get force required to push the block = 219.714 lb

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A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a

AVprozaik [17]

<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$ </h2>

Explanation:

A meteoroid is in a circular orbit 600 km above the surface of a distant planet.

Mass of the planet = mass of earth = 5.972 x $10^{24}$ Kg

Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km

The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = $\frac{GM}{r^{2} }$

where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x $10^{-11}$ $m^{3} kg^{-1} s^{-2}$

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

g = $\frac{(6.67 * 10^{-11}) (5.972 * 10^{24}) }{5733^{2} }$

g = 12.12 m/ $s^{2}$

The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$

6 0

3 years ago

A 91-kg astronaut and a 1300-kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, givi

WARRIOR [948]

Answer:

18.2145 meters

Explanation:

Using the conservation of momentum, we have that:

$m1v1 + m2v2 = m1'v1' + m2'v2'$

m1 = m1' is the mass of the astronaut, m2=m2' is the mass of the satellite, v1 and v2 are the inicial speed of the astronaut and the satellite (v1 = v2 = 0), and v1' and v2' are the final speed of the astronaut and the satellite. Then we have that:

$0 + 0 = 91*v1' + 1300*0.17$

$v1' = -1300*0.17/91 = -2.4286\ m/s$

The negative sign of this speed just indicates the direction the astronaut goes, which is the opposite direction of the satellite.

If the astronaut takes 7.5 seconds to come into contact with the shuttle, their initial distance is:

$distance = 2.4286 * 7.5 = 18.2145\ meters$

8 0

3 years ago

A two slit pattern is viewed on a screen 1.00m from the slits if the two third-order minima are 22.0 cm apart what is the width

Bingel [31]

Answer:

4.4 cm

Explanation:

Given:

Distance of the screen from the slit, D = 1 m

Distance between two third order interference minimas, x = 22 cm

Let's say, minima occurs at:

$x_n = (n + \frac{1}{2}) \frac{wL}{d}$

We have:

$2x_2 = 2(2 + \frac{1}{2}) * \frac{w*22}{d}$

Calculating further for the width of the central bright fringe, we have:

$\frac{w}{d} = \frac{22}{5}$

= 4.4 cm

Note: w in representswavelength

8 0

3 years ago

A crane lifts a 545 kg piano up into a high-rise apartment 75.0 meters above the ground. In order to do so, 5350 Newtons were ap

vodka [1.7K]

Answer:

Work done to pull the piano upwards is 401250 J

Explanation:

Work is done against the gravity to pull the piano upwards

So here we can say that work done is

$W = mgH$

here we know that

$mg = 5350 N$

also we know that

H = 75 m

now we have

$W = 5350 \times 75$

$W = 401250 J$

7 0

3 years ago

Which best explains how the diffraction pattern observed in young’s experiment supports the wave theory of light?.

KIM [24]

Answer: the airy pattern can only arise from wave propagation

Explanation:if particles went in straight lines through a slit, they would progate linearly and not interfere. The airy pattern arises from diffraction as waves interfere, producing peaks (constructive interference where peaks of waves from each slit coincide) and troughs (destructive interference where peaks and troughs of waves from each slit cancel out). If intensity rather than field is measured nodes occur where 0 values line up instead of troughs

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2 years ago