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solong [7]
3 years ago
5

A car travels 20km West and then 20 km South. What is the magnitude of it's displacement vector

Physics
1 answer:
Ludmilka [50]3 years ago
6 0

c. 28.km
11. Cartesian Coordinates.
X Y
S -3.5 4.5
T 0 -6.5
U 3.5 -2.5
what is angle of resultant vector STU measuring from positive x axis.
24 degrees below x

12. Cartesian Coordinates
Cartesian Coordinates.
X Y
S -3.5 4.5
T 0 -6.5
U 3.5 -2.5
what is magnitude of resultant vector STU
a. 4.9 b. 24 c. 16. d. 18
Seems right option is not there. The right one will be 7./2 ie 9.9
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A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
3 years ago
Una barra de plata de 335.2 g con una temperatura de 100 ºC se introduce un calorímetro de aluminio de 60 g de masa que contiene
sdas [7]

Respuesta:

0,0560 cal / gºC.

Explicación:

Cantidad de calor; (Q)

Q = mcΔt; Δt = t2 - t1

m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura

c de agua = 1 cal / gºC

c de aluminio = 0,22 cal / gºC

QTotal = Q de agua + Q de aluminio

Q de agua = 450 * 1 * (26 - 23) = 1350 cal

Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal

QTotal = 1350 + 39,6 = 1389,6 cal

Calor perdido = calor ganado

QTotal = calor perdido

- 1389,6 = 335,2 * c * (26 - 100)

-1389,6 = −24804,8 * c

c = 1389,6 / 24804,8

c = 0,056021 cal / gºC.

Capacidad calorífica específica de la plata = 0,0560 cal / gºC.

8 0
3 years ago
What did early experiments and Coulomb’s Law describe? Select all that apply.
Luda [366]
<span>Like charges repel and opposite charges attract.
The further away two charged objects are the weaker the electrical force between them.
The closer two charged objects are the stronger the electrical force between them.
Hope this helps :)</span>
4 0
3 years ago
Read 2 more answers
What distance does light travel in water, glass, and diamond during the time that it travels 1.0 m in vacuum? The refractive ind
lidiya [134]

Answer:

refractive index for water,glass,diamond are 0.752m, 0.667m, 0.413m respectively

Explanation:

refractive index (n) = \frac{velocity of light in air/vacuum}{velocity of light in substance}

velocity =\frac{distance}{time}

The time for travel is kept constant for all mediums.

refractive index (n) = \frac{\frac{distance in vacuum}{time} }{\frac{distance in medium}{time} }\\ \\=\frac{distance in vacuum}{distance in medium}

distance in medium = \frac{distance in vacuum}{refractive index of medium}

S_{medium}  = \frac{S_{vacuum} }{n_{medium} }

For water, n= 1.33

S_{water} = \frac{1 }{1.33}

S_{water} = 0.752m

For glass, n=1.5

S_{glass}= \frac{1 }{1.5}

S_{glass} = 0.667m

For diamond, n= 2.42

S_{diamond} = \frac{1 }{2.42}

S_{diamond} = 0.413m

3 0
3 years ago
The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3
viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;

\Rightarrow mg\sin{\theta} = F

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at \theta. Solving for the angle, we get

\sin{\theta} = \dfrac{F}{mg}

or

\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)

\;\;\;=  \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]

\;\;\;=42.9°

6 0
2 years ago
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