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stira [4]
3 years ago
14

A 31 kg crate full of very cute kittens is placed on an incline that is 17° below the horizontal. The crate is connected to a sp

ring that is anchored to a vertical wall, such that the spring is parallel to the surface of the incline.
a. If the crate was connected to the spring at equilibrium length, and then allowed to stretch the spring until the crate comes to rest, determine the spring constant. Assume that the incline is frictionless and that the change in length of the spring is 2.13 m.
b. If there is friction between the incline and the crate, would the spring stretch more, or less than if the incline is frictionless? You must use concepts pertaining to work and energy to receive full credit.
Physics
1 answer:
fgiga [73]3 years ago
5 0

Answer:

Explanation:

Change in length of spring = 2.13 m

Component of weight acting on spring = mg sinθ

so

mg sinθ = k x where k is spring constant and x is total stretch due to force on the spring.

Here x = 2.13

mg sin17 = k x 2.13

31 x 9.8 sin17 = k x 2.13

k = 41.7 N/m

b ) In case surface had friction , spring would have stretched by less distance .

It is so because , the work done by gravity in stretching down is stored as potential energy in  spring . In case of dissipative force like friction , it also takes up some energy in the form of heat etc  so spring stretches less.

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Leftover planetesimals that formed in the region of the solar system now occupied by the jovian planets are called
rewona [7]

Answer:

Comets

Explanation:

Comets are planetary celestial bodies consisting of ice and dust, sometimes rocky particles  formed in the region of the solar system. Long-period comets propagate towards the Sun by gravitational perturbations caused by passing stars. Some comets usually hyberbolic comets, move through the inner Solar System  prior to entering the interstellar region. Short period comet lies beyond the orbit of the Neptune.

The Jovian planets include Jupiter, Saturn, Uranus, and Neptune.

Therefore, leftovers of comets (planetesimal bodies) formed in the region of the solar system that are now occupied by the Jovian planets is due to the dusty particles  associated with the comets.

7 0
3 years ago
A car starts out traveling at 35 m/s. The car hits the brakes and decelerates at a rate of 3 m/s^2 for 5 seconds. What Distance
Ipatiy [6.2K]
Answer:

Time needed: 2.5 s
Distance covered: 31.3 m

Explanation:

I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by

v2f=v2i−2⋅a⋅d

Isolate d on one side of the equation and solve by plugging your values

d=v2i−v2f2a

d=(15.02−10.02)m2s−22⋅2.0ms−2

d=31.3 m

To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation

vf=vi−a⋅t, which will get you

t=vi−vfa

t=(15.0−10.0)ms2.0ms2=2.5 s

6 0
3 years ago
Calculate the required rate of return for Climax Inc., assuming that (1) investors expect a 4.0% rate of inflation in the future
Westkost [7]

Answer:

Required rate of return = 18.5 %

Explanation:

given,                            

rate of inflection = 4 %

risk free rate = 3 %                      

market risk premium = 5 %                    

firm has a beta  = 2.30                                  

rate of return has averaged 15.0% over the last 5 years

now,                                                                      

Nominal risk free rate = risk free rate  + inflation

                                    = 3%  +  4%

                                   = 7%

Required rate of return = Nominal risk free rate + β (RPM)

                                       = 7%       +          2.3 x 5.0%

Required rate of return = 18.5 %

5 0
3 years ago
A long copper rod of diameter 2.0 cm is initially at a uniform temperature of 100°C. It is now exposed to an air stream at 20°C
Lisa [10]

Answer:

t = 4.0 min

Explanation:

given data:

diameter of rod = 2 cm

T_1 = 100 degree celcius

Air stream temperature =  20 degree celcius

heat transfer coefficient = 200 W/m2. K

WE KNOW THAT

copper thermal conductivity = k = 401 W/m °C

copper specific heat Cp = 385 J/kg.°C

density of copper = 8933 kg/m3

charateristic length is given as Lc

Lc = \frac{V}{A_s}

Lc = \frac{\frac{\pi D^2}{4} L}{\pi DL}

Lc = \frac{D}{4}

Lc = \frac{0.02}{6} = 0.005 m

Biot number is given as Bi = \frac{hLc}{k}

Bi = \frac{200*0.005}{401}

Bi = 0.0025

As Bi is greater than 0.1 therefore lumped system analysis is applicable

so we have

\frac{T(t) - T_∞}{Ti - T_∞} = e^{-bt} ............1

where b is given as

b = \frac{ hA}{\rho Cp V}

b = \frac{ h}{\rho Cp Lc}

b = \frac{200}{8933*385*0.005}

b = 0.01163 s^{-1}

putting value in equation 1

\frac{25-20}{100-20} = e^{-0.01163t}

solving for t we get

t = 4.0 min

6 0
3 years ago
An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis
Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

W=\Delta K (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

W=\Delta K + E_{lost}

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane (\Delta K) and part is lost because of the air resistance (E_{lost}).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N

Now we can calculate the acceleration of the plane, by using Newton's second law:

a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

v^2- u^2 = 2aS

where

v=? is the final velocity

u=66.0 m/s is the initial velocity

a=1.69 m/s^2 is the acceleration

S=5.00 \cdot 10^2 m is the distance travelled

Solving for v, we find

v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s

8 0
3 years ago
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