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ivann1987 [24]
3 years ago
5

For physics show-and-tell, you want to impress the class by sipping water through a long straw. If you hold the straw vertically

, determine the length of the longest straw that you could, in principle, drink from in this manner._____ mAdditional Materials Show My Work (Optional)
Physics
1 answer:
yulyashka [42]3 years ago
8 0

Answer:

10.31 m

Explanation:

To find the result you have to use the formula: h = p / eg.

Where: e = 1000Kg/m^3          g=9.8m/s^2     p= 1.01 * 10^5 Pa

The calculation will be:

h= 1.01 * 10^5 / (1000kg/m^2) * (9.8m/s^2)

h= 10.31 m

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Item 19 Question 1 A Ferris wheel has a radius of 75 feet. You board a car at the bottom of the Ferris wheel, which is 10 feet a
sergey [27]

Answer:

y = 104.4 ft

Explanation:

As we know that we board in the car of ferris wheel at the bottom position

So we will have

final height of the car at angular displacement given as

y = y_o + R + R sin(270 - 255)

y = y_o + R + R sin 15

here we know that

y_o = 10 ft

R = 75 ft

so we have

y = 10 + 75 + 75 sin15

y = 104.4 ft

4 0
3 years ago
Estimate frequency of vibration of your arm. Let the length of the arm be 0.57 m. Consider the arm as a simple pendulum and assu
skad [1K]

Answer:

0.80865 Hz

1.23662 seconds

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

l = Length of arm = 0.57 m

Length of simple pendulum is given by

L=\dfrac{2}{3}l\\\Rightarrow L=\dfrac{2}{3}\times 0.57\\\Rightarrow L=0.38\ m

The frequency is given by

f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{9.81}{0.38}}\\\Rightarrow f=0.80865\ Hz

The frequency is 0.80865 Hz

The time period is given by

T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{0.80865}\\\Rightarrow T=1.23662\ s

The time period is 1.23662 seconds

3 0
3 years ago
What is the least force you could exert on the refrigerator to move it?
Ulleksa [173]
 <span>If the refrigerator weights 1365 and you are not exerting any vertical force on it, then the normal force is also 1365N. so Fn=1365 

Fsf = Static frictional force = (coefficient of static friction) * (Normal force) 

So the least for you could exert to move it is equal to the Fsf. 
Fsf = (0.49)(1365N)</span><span>
</span>
8 0
3 years ago
A particle with a charge of − 5.10 nC is moving in a uniform magnetic field of B⃗ =−( 1.20 T )k^. The magnetic force on the part
marta [7]

Answer:

Explanation:

Given that,

Charge q=-5.10nC

Magnetic field B= -1.2T k

And the magnetic force

F =−( 3.30×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

F= q(v×B)

−( 3.30×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.2k

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2x i×k - 1.2y j×k - 1.2z k×k)

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( 1.2xj - 1.2y i )

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2y i + 1.2x j)

So comparing comparing coefficients

let compare x axis component

-( 3.30×10−7N )i=-1.2qy i

−3.30×10−7N = -1.2qy

y= -3.3×10^-7/-1.2q

y= -3.3×10^-7/-1.2×-5.10×10^-9)

y=-53.92m/s

Let compare y-axisaxis

7.6×10−7N j = 1.2qx j

7.6×10−7N = 1.2qx

x= 7.6×10^-7/-1.2q

x= 7.6×10^-7/1.2×-5.10×10^-9)

x=-124.18m/s

a. Then, the velocity of the x component is x= -124.18m/s

b. Also, the velocity component of the y axis is =-53.92m/s

c. We will compute

V•F

V=-124.18i -53.92j

F=−( 3.30×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

V•F is

(-124.18i-53.92j)•−(3.30×10−7N)i+(7.60×10−7 N )j =

4.1×10^-5 - 4.1×10^-5=0

V•F=0

d. Angle between V and F

V•F=|V||F|Cosx

0=|V||F|Cos

Cosx=0

x= arccos(0)

x=90°

Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

6 0
3 years ago
child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s
Strike441 [17]

Answer:

θ = 13.7º

Explanation:

  • According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
  • The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
  • As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

       \Delta K + \Delta U = W_{nc} (1)

  • ΔK, is the change in kinetic energy, as follows:

       \Delta K = \frac{1}{2}* m* (v_{f} ^{2}  - v_{0} ^{2}) (2)

  • ΔU, is the change in the gravitational potential energy.
  • If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

        \Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)

  • Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

        W_{nc} = F_{f} * d * cos 180\º \\\\  = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)

  • Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

      \frac{1}{2}* (v_{f} ^{2}  - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d

  • Replacing by the givens and the knowns, we can solve for sin θ, as follows:              \frac{1}{2}*( (4.30 m/s) ^{2}  - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2}  = 0.236⇒ θ = sin⁻¹ (0.236) = 13.7º
8 0
3 years ago
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