All categories

3 years ago

In a uniform circular motion map, what is always true? Check all that apply.

Physics

2 answers:

Katen [24]3 years ago

7 0

Answer:

Acceleration vectors point toward the center of the circle.

Velocity vectors are the same length.

Explanation:

grandymaker [24]3 years ago

6 0

Answer:

Velocity vectors are always perpendicular to the circle.

Acceleration vectors point toward the center of the circle.

Velocity vectors are the same length.

Explanation:

(THESE ARE NOT MY WORDS BTW)

1) Acceleration and velocity are vectorial quantities, which means they have magnitude and direction.

2) In a circular motion velocity direction changes all the time, which means that it is accelerated.

3) In a uniform circular motiion, the velocity changes in a constant value. This is the rate of change of velocity, which is the magnitude of the acceleration, is constant (uniform).

4) The velocity is perpendicular to the path, i.e. the circle. You can see it if you think that if the object stopped changing the direction, then the object would follow a straight path (as per inertia principle). That is why this velocity is called tangential velocity (to differentiate it of the angular velocity).

This is what the option C says "Velocity vectors are always perpendicular to the circle". Then this is true.

5) The constant change of direction in a circular path, means that the object is been pushed, accelerated, toward the center of a circle. This is, all the time the object in motion tries to follow the perpendicular path but a push (a force) directed to the center of the circle changes its direction. Such force accelerates the object toward the center of the circle. So, the acceleration vectors point toward the center of the circle, which is what the option D says. So, this is also true.

6) Since the motion is uniform, the magnitude or length of the velocity vectors are always the same, are constant. So, the option E. is also true.

You might be interested in

Por una tubería de 0.06 m de diámetro circula agua con una velocidad desconocida, al llegar a la parte estrecha de la tubería de

Vesnalui [34]

Answer:

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 $\frac{m}{s}$

Explanation:

La ecuación de continuidad es simplemente una expresión matemática del principio de conservación de la masa. Este principio establece que la masa de un objeto o colección de objetos nunca cambia con el tiempo.

La ecuación de continuidad es la relación que existe entre el área y la velocidad que tiene un fluido en un lugar determinado y dice que el caudal de un fluido es constante a lo largo de un circuito hidráulico.

En otras palabras, la ecuación de continuidad se basa en que el caudal (Q) del fluido ha de permanecer constante a lo largo de toda la conducción. Cuando un fluido fluye por un conducto de diámetro variable, su velocidad cambia debido a que la sección transversal varía de una sección del conducto a otra.

Entonces, siendo el caudal es el producto de la superficie de una sección del conducto por la velocidad con que fluye el fluido, en dos puntos de una misma tubería se cumple:

Q1=Q2

A1*v1= A2*v2

donde:

A es la superficie de las secciones transversales de los puntos 1 y 2 del conducto.

v es la velocidad del flujo en los puntos 1 y 2 de la tubería.

Siendo $A=pi*r^{2} =pi*(\frac{D}{2} )^{2} =\frac{pi*D^{2} }{4}$ , donde pi es el número π, r es el radio del conducto y D el diámetro del conducto, entonces:

$\frac{pi*D1^{2} }{4}*v1=\frac{pi*D2^{2} }{4}*v2$

En este caso:

D1: 0.06 m
v1: ?
D2: 0.04 m
v2: 2.6 m/s

Reemplazando:

$\frac{pi*(0.06m)^{2} }{4}*v1=\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}$

Resolviendo:

$v1=\frac{\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}}{\frac{pi*(0.06m)^{2} }{4}}$

$v1=\frac{(0.04m)^{2} }{(0.06m)^{2} }*2.6\frac{m}{s}$

v1= 1.156 $\frac{m}{s}$

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 $\frac{m}{s}$

8 0

3 years ago

A spring has a constant k= 10.32 N/m and is hung vertically. A 0.400kg mass is suspended from the spring. What is the displaceme

alukav5142 [94]

Answer:

displacement (x) = 0.003798 meters

Explanation:

from the fact that the string is hung vertically we can deduce that:

Total force acting on the mass = Fs (by spring) + Fg (by gravity)

where

Fs = k*x , x is the displacement..

Fg = m*g

then:

Ftot = m*a, but a = 0 m/(s^2) because the mass becames stationary.

Ftot = 0

Fs + Fg = 0

by direction, take down as negative.

Fs - Fg = 0

k*x = m*g

x = m*g/k = [(0.400)(9.8)]/(10.32)

= 0.3798 meters

4 0

3 years ago

A magnet of mass 0.20 kg is dropped from rest and falls vertically through a 35.0 cm copper tube. Eddy currents are induced, cau

topjm [15]

Answer:

0.58 J

Explanation:

We know that Total energy is conserved.

Initial Kinetic energy + Initial potential energy = final kinetic energy+ final potential energy + dissipated heat energy

Initial kinetic energy = 0 ( magnet is at rest initially)

Initial Potential energy = m g h = (0.20 kg)(9.81 m/s²)(0.35 m) = 0.69 J

Final kinetic energy = 0.5 m v² = 0.5 ×0.20 kg × 1.10 m/s = 0.11 J

Final potential energy = 0

∴ Dissipated heat energy = (0.69 -0.11) J = 0.58 J

4 0

3 years ago

A compound is found to have a molar mass of 598 g/mol. if 0.0358 g of the compound is dissolved in enough water to make 175 ml o

zalisa [80]

π=iMRT

Where, π is Osmotic pressure,
i=1 for non-electrolytes,
M is molar concentration of dissolved species (units of mol/L)
R is the ideal gas constant = 0.08206 L atm mol⁻¹K⁻¹,
T is the temperature in Kelvin(K),

Here, to calculate M convert into standard units mg tog, ml to L, c to Kelvin
M= ( $\frac{35.8}{598}$ *10⁻³ )/ 0.175 =(5.987 *10⁻⁵)mol / 0.175L = 34.21*10⁻⁵ mol/L

π=iMRT=(1)*(34.21*10⁻⁵)*(0.08206)*(298.15)=837×10⁻⁵= 8.37×10⁻³ atm
=6.36 torr
(1 atm=760 torr, 1 Kelvin =273.15 °C, 1L=1000ml, 1g=1000mg)

3 0

3 years ago

What is a factor that limits a technological design?

Ivanshal [37]

The answer is a constraint

5 0

4 years ago

Read 2 more answers