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3 years ago

13

Please help me! It states constant speed but moving in circular motion. I know there is acceleration due to a rate of change of

velocity but what is the direction

1 answer:

Morgarella [4.7K]3 years ago

8 0

Direction of resultant force in circular motion is directed towards the center of the circle. Hence vector B is the direction of the resultant force.

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A bicyclist starting at rest produces a constant angular acceleration of 1.30 rad/s2 for wheels that are 35.5 cm in radius.

Debora [2.8K]

Answer:

a) 0.462 m/s^2

b) 31.5 rad/s

c) 381 rad

d) 135m

Explanation:

the linear acceleration is given by:

$a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2$

the angular speed is given by:

$\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s$

to calculate how many radians have the wheel turned we need the apply the following formula:

$\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad$

the distance is given by:

$d=\theta*r$

$d=381rad*(35.5*10^{-2}m)\\d=135m$

4 0

3 years ago

When water moves from a gas state to a liquid state, it is experiencing ______.

son4ous [18]

It is called condensation. Hope this helped!

8 0

2 years ago

Read 2 more answers

*How much energy is<br>transferred in lifting a 5 kg<br>Mass 3m

AlexFokin [52]

Answer:

147 J

Explanation:

The energy transferred to potential energy is :

U = m * g * h = (5 kg) * (9.8 m/s^2) * (3 m) = 147 J

3 0

3 years ago

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

3 years ago

A bubble of air is rising up through the ocean. When it is at a depth of 20.0 m below the surface, where the temperature is 5.00

kotegsom [21]

Answer:

the volume is 0.253 cm³

Explanation:

The pressure underwater is related with the pressure in the surface through Pascal's law:

P(h)= Po + ρgh

where Po= pressure at a depth h under the surface (we assume = 1atm=101325 Pa) , ρ= density of water ,g= gravity , h= depth at h meters)

replacing values

P(h)= Po + ρgh = 101325 Pa + 1025 Kg/m³ * 9.8 m/s² * 20 m = 302225 Pa

Also assuming that the bubble behaves as an ideal gas

PV=nRT

where

P= absolute pressure, V= gas volume ,n= number of moles of gas, R= ideal gas constant , T= absolute temperature

therefore assuming that the mass of the bubble is the same ( it does not absorb other bubbles, divides into smaller ones or allow significant diffusion over its surface) we have

at the surface) PoVo=nRTo

at the depth h) PV=nRT

dividing both equations

(P/Po)(V/Vo)=(T/To)

or

V=Vo*(Po/P)(T/To) = 0.80 cm³ * (101325 Pa/302225 Pa)*(277K/293K) = 0.253 cm³

V = 0.253 cm³

3 0

3 years ago