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2 years ago

State a true conclusion.

Physics

1 answer:

Ganezh [65]2 years ago

7 0

The conclusion is; Cathy's dad has gone fishing.

<h3>What are conclusions in conditional statements?</h3>

A conditional statement is a statement with a hypothesis which is then followed by a conclusion.

A conditional statement has two parts; "if" and "then"

The hypothesis is “if,” part of a conditional statement.

The conclusion is the “then,” part of a conditional statement.

Considering the given conditional statements:

1. If it's Saturday, then Cathy's dad goes fishing.

2. If it's Saturday, then Cathy's dad goes fishing.

Since the if part is true, the conclusion will be, Cathy's dad has gone fishing.

In conclusion, a conditional statement consists of a hypothesis and a conclusion.

Learn more about conditional statements at: brainly.com/question/21170
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The unit of area is a derived unit. Why?

prohojiy [21]

Explanation:

area=length(m) ×breadth(m) . The unit of area is expressed in terms of fundamental units m^2.thus it is derived unit

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3 years ago

A ball is thrown vertically upward from the top of a building 80 feet tall with an initial velocity of 64 feet per second. The d

-BARSIC- [3]

Answer:

a) $t=6.37s$

b) $t=3.3333s$

Explanation:

The knowable variables are the initial hight and initial velocity

$s_{o}=80ft$

$v_{os}=64ft/s$

The equation that describes the motion of the ball is:

$s=80+64t-16t^{2}$

If we want to know the time that takes the ball to hit the ground, we need to calculate it by doing s=0 that is the final hight.

$0=80+64t-12t^{2}$

a) Solving for t, we are going to have two answers

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

a=-16

b=64

c=80

$t=-1.045 s$ or $t=6.378s$

<em><u>Since time can not be negative the answer is t=6.378s </u></em>

b) To find the time that takes the ball to pass the top of the building on its way down, we must find how much does it move too

First of all, we need to find the maximum hight and how much time does it take to reach it:

$v_{y}=v_{o}+gt$

at maximum point the velocity is 0

$0=64-32.2t$

Solving for t

$t=1.9875 s$

Now, we must know how much distance does it take to reach maximum point

$s=0+64t-16t^{2} =64(1.9875)-12(1.9875)^{2} =80ft$

So, the ball pass the top of the building on its way down at 160 ft

$160=80+64t-16t^{2}$

Solving for t

$t=2s$ or $t=3.333s$

Since the time that the ball reaches maximum point is almost t=2s that answer can not be possible, so the answer is t=3.333s for the ball to go up and down, passing the top of the building

4 0

3 years ago

Read 2 more answers

if a string of light goes out when one of the bulbs is removed, are the lights probably connected in a series circuit or a paral

Minchanka [31]

Yes it is because other wise the light would stay on

4 0

3 years ago

Read 2 more answers

Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second

Pie

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the <u>paths</u> meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .

Setting the locations equal, we get the following equations to solve for $T_F$ and $T_S$ :

$(-1 + T_F) = (-7 + 2T_S)$ Equation 1

$(4 - T_F) = (-6 + 2T_S)$ Equation 2

$(-1 + 2T_F) = (-1 + T_S)$ Equation 3

Solving these three equations simultaneously we get:

$T_F =$ 2 seconds

$T_S =$ 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of $T_F$ or $T_S$ in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

5 0

3 years ago

A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the

kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂ : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F : Force ( N)

d : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the bar

mm = 1.5 kg : mass of the monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances from forces the point 2

d₁₂ = (3-0.6) m = 2.4m : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium of moments at the point 2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0

3 years ago