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s344n2d4d5 [400]
3 years ago
5

For a block to move down an inclined plane what force has to be the greatest?

Physics
1 answer:
Hatshy [7]3 years ago
5 0

Answer:

D) True. This is what creates the body weight

Explanation:

Let's write Newton's second law for this case. For inclined planes the reference system takes one axis parallel to the plane (x axis) and the other perpendicular to the plane (y axis)

X axis

          Wx -fr = ma

Y Axis

          N - Wy = 0

With trigonometry we can find the components of weight

          sin θ = Wₓ / W

         cos θ = W_{y} / W

         Wₓ = W sin θ

          W_{y} = W cos θ

        W  sin θ - fr = ma

From this expression as it indicates that the body is descending the force greater is the gravity that create the weight of the body

Let's examine the answers

A False This force does not apply because it is not a spring

B) False. It is balanced at all times with the component (Wy) of the weight

C) False. For there to be a rope, if it exists you should be less than the weight component for the block to lower

D) True. This is what creates the body weight

E) False. The cutting force occurs for force applied at a single point and gravity is applied at all points

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A solenoid used to produce magnetic fields for research purposes is 2.1 m long, with an inner radius of 28 cm and 1000 turns of
Veseljchak [2.6K]

Answer:

I = 2172.46 A

Explanation:

Given that,

The length of a solenoid, l = 2.1 m

The inner radius of the solenoid, r = 28 cm = 0.28 m

The number of turns in the wire, N = 1000

The magnetic field in the solenoid, B = 1.3 T

We need to find the current carried by it. We know that, the magnetic field in a solenoid is given by :

B=\mu_o nI\\\\or\\\\B=\mu_o \dfrac{N}{L}I\\\\I=\dfrac{BL}{\mu_o N}

Put all the values,

I=\dfrac{1.3\times 2.1}{4\pi \times 10^{-7}\times 1000}\\\\I=2172.46\ A

So, it carry current of 2172.46 A.

7 0
3 years ago
For a maximum superelevation of 0.08 ft/ft and a degree of curve of 4o, calculate the maximum safe speed for the curve assuming
Mamont248 [21]

Answer:

Explanation:

Given that

Superelation= 0.08ft/ft

Given curve= u•

Curve junction factor= 0.13

DR= 5729.57795

R = 5729.57795/D

R = 5729.57795/4

R = 1432.4ft

c + f = V^2/gG

0.08 + 0.13 = V^2 / (32*1432.4)

V^2 = 9625.728 or V = 98 ft/sec

The designed speed for a project considered is a minimum value which means the highway design elements will meet or exceed the standards for the design speed. The maximum safe speed under normal condition is significantly greater than design speed

7 0
3 years ago
A 535 kg roller coaster car began at rest at the top of a 93.0 m hill. Now it is at the top of the first loop-de-loop.
iVinArrow [24]
Using g = 9.8 m/s2, the statement that best describes the roller coaster car when it is at the top of the loop-de-loop is that The car has both potential and kinetic energy, and it is moving at 24.6 m/s. The correct answer is <span>B) The car has both potential and kinetic energy, and it is moving at 24.6 m/s.</span>
4 0
3 years ago
Read 2 more answers
Calculate the volume of the metal required to make a hemisperical bowl with internal and external radii 8.4cm and 9.1cm respecti
ioda
External = R
Internal = r
Volume of hemisperical = 2/3 π(R³-r³)
V= 2/3 π(9.1³ - 8.4³)
V= 336.9 cm³
4 0
2 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
3 years ago
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