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Travka [436]
3 years ago
5

A 244.0 N block is at rest on a flat, frictionless table. A hooked cable applies an upward force of 24.0 N on the block. What is

the normal force acting on the block?
Physics
1 answer:
blagie [28]3 years ago
4 0

Answer:

268N

Explanation:

The upward force acting on the block are the reaction and the hooked table..

The total normal force acting = normal reaction + 24N

Note that the normal reaction is always equal the weight of the table

Hence the normal force acting in the block is 244.0+24 = 268.0N

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If you see a sedimentary rock outcrop and red layers of sand are on top of pale yellow layers of sand, what do you know for sure
neonofarm [45]

Answer: The yellow layer is definitely older than the red layer

Explanation: According to Nicolaus Steno's law of superposition and original horizontality. Older rocks underlie younger rocks.

Sedimentary rocks are usually deposited in horizontal layers in which each stratigraphic layer is laid down before another can be deposited upon it.

The red layer, in addition to being older, is also likely to have undergone intense oxidation due to earlier exposure.

5 0
3 years ago
A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it
Darina [25.2K]

Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

<u>Given:  </u>

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.  

<u>Solution  </u>

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next  

Ef=Ei+W                                                 (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).  

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form  

(Kf + mc^2) = (KJ+ mc^2)                       (2)  

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.  

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W                              (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by  

W = FΔr                                                      (4)  

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut  

W = F Δr= (190N)(6 m) = 1140 J  

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate  

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

This is the final speed of the first push when the astronaut applies a positive work done. Then, in the second push, he applies a negative work done on the crate to slow down its speed. Hence, in this case, we could consider the initial speed of the second process to be the final speed of the first process. So,  

vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W'                                 (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by  

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate  

1/2mvf'^2=1/2mvi'^2+W'

  vf'=4.50 m/s

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

7 0
3 years ago
A 110 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. How much work must be d
Softa [21]

Answer:

the work that must be done to stop the hoop is 2.662 J

Explanation:

Given;

mass of the hoop, m = 110 kg

speed of the center mass, v = 0.22 m/s

The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;

W = ΔK.E

W = ¹/₂mv²

W = ¹/₂ x 110 x 0.22²

W = 2.662 J

Therefore, the work that must be done to stop the hoop is 2.662 J

6 0
3 years ago
A block slides down a frictionless inclined ramp. If the ramp angle is 17.0° and its length is 30.0 m, find the speed of the blo
shutvik [7]

Answer:

 v = 17.15 m/s

Explanation:

given,

angle of ramp = 17.0°

length of ramp(l) = 30 m

height of the ramp =  

     h = l sin \theta

     h = 30 sin 30^0

            h = 15 m

using energy of conservation

\dfrac{1}{2}mv^2 = mgh

\dfrac{1}{2}v^2 = gh

v = \sqrt{2gh}

v = \sqrt{2\times 9.8\times 15}

v = \sqrt{294}

 v = 17.15 m/s

 speed of block reaching at the bottom = v = 17.15 m/s

5 0
3 years ago
A diffraction grating produces a first-order bright fringe that is 0.18 m away from the central bright fringe on a flat screen.
Georgia [21]

Answer:

The wavelength of the light is 562.5 nm

Solution:

As per the question:

Order, n = 1

Slit separation, d = 2.5\times 10^{- 6}\ m

Distance from the bright fringe, y = 0.18 m

Distance between the screen and the grating, D = 0.8 m

Now,

We know from the eqn for diffraction:

n\lambda = dsin\theta

n = 1

\lambda = dsin\theta            (1)

Also,

For very small angle, \theta:

sin\theta ≈ tan\theta = \frac{y}{D} = \frac{0.18}{0.8} = 0.225

Using the above value in eqn (1):

\lambda = 2.5\times 10^{- 6}\times 0.225 = 5.625\times 10^{- 7}\ m = 562.5\ nm

3 0
3 years ago
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