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brilliants [131]
3 years ago
14

What is a centripetal acceleration of a point on a bicycle wheel of a radius of 0.70 m when a bike is moving 8.0 m/s

Physics
1 answer:
Furkat [3]3 years ago
3 0

Answer:

The acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

Explanation:

The centripetal acceleration acts on a body when it is performing a circular motion.

Here, a point on the bicycle is performing circular motion as the rotation of the wheel produces a circular motion.

The centripetal acceleration of a point moving with a velocity v and at a distance of r from the axis of rotation is given as:

a=\frac{v^2}{r}

Here, v=8\ m/s,r=0.70\ m

∴ a=\frac{8}{0.70}=11.43\ m/s^2

Therefore, the acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

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A rock is thrown at a window that is located 18.0 m above the ground. The rock is thrown at an angle of 40.0° above horizontal.
Korvikt [17]

Answer:

B) 27.3 m

Explanation:

The rock describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x =  vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

(vfy)² = (v₀y)² - 2g(y- y₀)    Equation (2)

vfy = v₀y -gt    Equation (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 30 m/s , at an angle  α=40.0° above the horizontal

v₀x = vx = 30*cos40° = 22.98 m/s

v₀y = 30*sin40° = 19.28 m/s

y₀ = 2m

y =  18.0 m

g = 9.8 m/s²

Calculation of the time (t) it takes for the rock to reach at  18 m above the ground

We replace data in the equation (2)

(vfy)² = (v₀y)² - 2g(y- y₀)    

(vfy)² = (19.28)² - 2(9.8)(18- 2)

(vfy)² = 371.86 - 313.6

(vfy)² = 58.26

v_{f} = \sqrt{58.26}

vfy = 7.63 m/s

We replace vfy = 7.63 m/s in the equation (2)

vfy = v₀y - gt

7.63 = 19.28 - (9.8)(t)

(9.8)(t) = 11.65

t = 11.65 / (9.8)

t = 1.19 s

Horizontal distance from where the rock was thrown to the window

We replace t = 1.19 s , in the equation (1)

x =  vx*t  

x = (22.98)* ( 1.19 )

x = 27.3 m

3 0
3 years ago
The nebular theory also predicts that the cloud should heat up as it collapses. what physical law explains why it heats up?
Anna [14]
The physical law that explains that is the law of conservation of energy which states that he energy of an isolated sistem remains constant
4 0
3 years ago
Give a specific example of a machine,and describe how its mechanical efficiency might be calculated
natta225 [31]
A machine that is interesting will be the homework machine. The way it works is you put your homework inside a slot, and you have to write any three letter word so the machine knows what handwriting to use. In 2 minutes your homework will be complete in your writing.
3 0
3 years ago
Imagine a 15 kg block moving with a velocity of 20 m/s to the left. Calculate the kinetic Energy of this block.
ivann1987 [24]

Answer:

3000 J

Explanation:

Kinetic energy is:

KE = ½ mv²

If m = 15 kg and v = -20 m/s:

KE = ½ (15 kg) (-20 m/s)²

KE = 3000 J

3 0
3 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Phoenix [80]

Answer:

The temperature of the core raises by 2.8^{o}C every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is 1.60\times 10^{5}kg will require

0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s

5 0
3 years ago
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