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Wewaii [24]
3 years ago
5

Your physics textbook is sliding to the right across the table draw the vectors starting at the black dot. the location and orie

ntation of the vectors will be graded. the exact length of your vectors will not be graded but the relative length of one to the other will be graded.
Physics
1 answer:
liberstina [14]3 years ago
6 0
You would take a black dot at the top right of the y axis and dray it to the the far right of your x axis. hope this helps have a nice day and God bless
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A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.
Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

We apply the Formula (1)

vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t   Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0
3 years ago
Acceleration is defined as the change in velocity divided by
mihalych1998 [28]

Answer:

change in time

Explanation:

8 0
3 years ago
Can somebody help me on finding the constants if this experiment?
ki77a [65]
The constants could be the day because the experiment is all on the same day or the person throwing the pumpkin because it will always be the same perosn
8 0
3 years ago
A swimmer is swimming to the left with a speed of 1.0\,\dfrac{\text m}{\text s}1.0 s m ​ 1, point, 0, start fraction, start text
bulgar [2K]

Answer:

2.9s

Explanation:

3 0
3 years ago
Read 2 more answers
What are the signs of the velocity, force, and acceleration (components) afterthe cart is released and is moving qway from the m
SVETLANKA909090 [29]

Answer:

car is moving away its direction is negative.

the speed must also be negative.

speed this distance decreases the acceleration is negative and if the speed is increasing the acceleration is positive.

Explanation:

In the exercise they indicate that the direction to the motion sensor is positive, as they indicate that the car is moving away its direction is negative.

The speed of the car is

           v = (x₂-x₁) / t

As the positions are negative, and the car moves away the speed must also be negative.

The analysis for acceleration must be very careful if the speed this distance decreases the acceleration is negative and if the speed is increasing the acceleration is positive.

3 0
3 years ago
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