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Vsevolod [243]
3 years ago
11

Anderson found the answers to his math assignment online and used them to finish his homework. When he was done he posted a stat

us update, "Homework finished. Thanks Yahoo! answers." Anderson's post will most likely encourage his friends to finish their homework on time provide evidence against him in an academic investigation prove he understands how to use technology effectively help other students find useful homework resources
Physics
1 answer:
Sonja [21]3 years ago
4 0

Answer:

provide evidence against him in an academic investigation

Explanation:

You might be interested in
A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r
Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

(c) at = 4 m/s²  ,  v = 42 m/s

Explanation:

The uniformly accelerated circular movement,  is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t  Formula (1)

θ=  ω₀*t + (1/2)*α*t²  Formula (2)

at = α*R  Formula (3)

v= ω*R  Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s  : wheel’s initial angular velocity

R = 1.0 m  : wheel’s radium

(a)  Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ=  ω₀*t + (1/2)*α*t²

θ=  (2)*(10) + (1/2)*(4)*(10)²

θ=  220 rad  

θ=  220 rad  

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0
3 years ago
He formula p=m x v to answer th<br>2,000 kg car moving at 30 m/s​
aivan3 [116]

The momentum of the car is 60,000 kg m/s

Explanation:

The momentum of an object is a vector quantity, whose magnitude is given by

p=mv

where

m is the mass of the object

v is its velocity

For the car in this problem, we have

m = 2,000 kg

v = 30 m/s

Therefore, the magnitude of its momentum is

p=(2000)(30)=60,000 kg m/s

Learn more about momentum here:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

7 0
3 years ago
Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their
geniusboy [140]

Answer:

a. 11 m/s at 76° with respect to the original direction of the lighter car.

Explanation:

In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

m_{1} v_{x1}+m_{2}v_{x2}=(m_{1}+m_{2})v_{fx}\\m_{1} v_{y1}+m_{2}v_{y2} =(m_{1}+m_{2})v_{fy}

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

1*13 + 4*0 = (1+4) v_{fx}\\v_{fx}=\frac{13}{5} =2.6\\1*0 + 4*13 = (1+4) v_{fy}\\v_{fy}=\frac{13*4}{5} =10.4\\

The magnitude of the final velocity of the wreck can be found as:

v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72

The final velocity has an intensity of roughly 11 m/s

As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

\theta = cos^{-1}(\frac{v_{fx} }{v_{f}} )\\\theta = cos^{-1}(\frac{2.6}{10.7} )\\\theta = 76^{o}

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.

5 0
3 years ago
A motorcycle accelerates from a stop at a rate of 4m/s 2 for 20 seconds. It then continues at a
8_murik_8 [283]

After 20 s, the motorcycle attains a speed of

\left(4\dfrac{\rm m}{\mathrm s^2}\right)(20\,\mathrm s)=80\dfrac{\rm m}{\rm s}

and it continues at this speed for the next 40 s. So at 45 s, its speed is 80 m/s.

4 0
3 years ago
What would you expect to cause a drop in air pressure?
Murrr4er [49]

Answer:

D

Explanation:

6 0
3 years ago
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