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artcher [175]
3 years ago
6

An earthquake causes a 3 kg book to fall from a shelf. If the book lands with

Physics
1 answer:
Anna [14]3 years ago
7 0
The correct answer is C
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The numbers of times you do lift without resting during a muscle training workout is called
erma4kov [3.2K]
1. Improves health
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4. Requires minimal equipment
5. Convenient
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3 0
3 years ago
Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li
Fofino [41]

Answer:

(a) t=3.87 s :time at which Kathy overtakes Stan

(b) d=37.36 m

(c) vf₁ = 15.097 m/s : Stan's final speed

    vf₂ = 19.31 m/s : Kathy's final speed

Explanation:

kinematic analysis

Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

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/s²

Nomenclature

d₁: Stan displacement   

t₁ :  Stan time

v₀₁: Stan initial speed

vf₁: Stan final speed

a₁:  Stan acceleration

d₂: car displacement   

t₂ : Kathy time

v₀₂: Kathy initial speed

vf₂: Kathy final speed

a₂:  Kathy acceleration

Data

v₀₁ = 0

v₀₂ = 0

a₁ = 3.1 m/s²

a₂= 4.99 m/s²

t₁ = (t₂ +1) s

Problem development

By the time Kathy overtakes Stan, the two will have traveled the same distance:

d₁ = d₂

t₁ = (t₂ +1)

We aplpy the Formula (3)

d₁ = v₀₁t₁ + (1/2)*a₁*t₁²

d₁ = 0 + (1/2)*(3.1)*t₁²

d₁ =  1.55*t₁² ; Stan's cinematic equation 1

d₂ = v₀₂t₂ + (1/2)*a₂*t₂²

d₂ = 0 + (1/2)*(4.99)*t₂²

d₂ = 2.495* t₂² : Kathy's cinematic equation 2

d₁ = d₂

equation 1 = equation 2

1.55*t₁²  =  2.495* t₂²  , We replace t₁ = (t₂ +1)

1.55* (t₂ +1) ² = 2.495* t₂²

1.55* (t₂² +2t₂+1) = 2.495* t₂²

1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²

1.55t₂²+3.1t₂+1.55=2.495t₂²

(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0

0.905t₂² - 3.1t₂ - 1.55 = 0  Quadratic equation

Solving the quadratic equation we have:

(a) t₂ = 3.87 s : time at which Kathy overtakes Stan

(b) Distance in which Kathy catches Stan

we replace t₂ = 3.87 s in equation 2

d₂ = 2.495*( 3.87)²

d₂ = 37.36 m

(c) Speeds of both cars at the instant  Kathy overtakes Stan

We apply the Formula (1)

vf₁= v₀₁+a₁t₁    t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s

vf₁= 0+3.1* 4.87

vf₁ = 15.097 m/s : Stan's final speed

vf₂ = v₀₂+a₂ t₂  

vf₂ =0+4.99* 3.87

vf₂ = 19.31 m/s : Kathy's final speed

8 0
3 years ago
Exit When an object is charged either positively or negatively, what can be said about the electric field surrounding the object
taurus [48]
Believe the ANWSER is b or a
7 0
3 years ago
A steel ball is dropped from a building's roof and passes a window, taking 0.115 s to fall from the top to the bottom of the win
marta [7]

Answer:

The building is 26.85m tall.

Explanation:

In order to solve this problem, we must start by drawing a diagram of the situation. (See attached picture).

We split the height of the building into three parts: y_{1}, the height of the window and y_2

In order to find each of those, we need to start by finding the velocities of the ball in points A and B. We will analyze the trajectory of the ball when bouncing back to the top of the building. Let's start by finding the velocity of the ball in A:

We can use the following formula to determine the velocity of the ball in part A:

\Delta y=V_{A}t+\frac{1}{2}at^{2}

which can be solved for V_{A}

so we get:

V_{A}=\frac{\Delta y-\frac{1}{2}at^{2}}{t}

and now we can substitute (remember the acceleration of gravity goes downward so we will consider it to be negative).

V_{A}=\frac{(1.30m)-\frac{1}{2}(-9.8m/s^{2})(0.115s)^{2}}{0.115s}

which yields:

V_{A}=11.87m/s

once we got the velocity at point A, we can now find the velocity at point B. We can do so by using the following formula:

a=\frac{V_{A}-V_{B}}{t}

which can be solved for V_{B} which yields:

V_{B}=V_{A}-at

so we can substitute values now:

V_{B}=11.87m/s-(-9.8)(0.115s)

Which yields:

V_{B}=13m/s

Now that we have the velocities at A and B, we can use them to find the values of y_{1} and y_{2}

Let's start with  y_{1}

We can use the following formula to find it:

y_{1}=\frac{V_{f}^{2}-V_{A}^{2}}{2a}

we know the final velocity of the rebound will be zero, so we can simplify our formula:

y_{1}=\frac{-V_{A}^{2}}{2a}

so we can substitute now:

y_{1}=\frac{-(11.87m/s)^{2}}{2(-9.8m/s^{2})}

which solves to:

y_{1}=7.19m

Now we can proceed and find the value of y_{2}

the value of y_{2} can be found by using the following formula:

y_{2}=V_{B}t-\frac{1}{2}at^{2}

in this case our t will be half of the tie spent below the bottom of the window, so:

t=\frac{2.04s}{2}=1.02s

so now we can substitute all the values in the given formula:

y_{2}=(13m/s)(1.02s)-\frac{1}{2}(-9.8m/s^{2})(1.02s)^{2}

which yields:

y_{2}=18.36m

so now that we have all the values we need, we can go ahead and calculate the height of the building:

h=y_{1}+window+y_{2}

when substituting we get:

h=7.19m+1.3m+18.36m

So the answer is:

h=26.85m

The building is 26.85m tall.

5 0
3 years ago
I will give anyone who answers this fast and correctly a brainliest on this and go to your account and give 5 stars/thanks on al
kykrilka [37]

Answer:I will give it to another person

Explanation:because they are good to me

3 0
2 years ago
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