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Anika [276]
3 years ago
9

Consider the following hypotheses: 12:V ) >(x)) Use rules of inference to prove that the following conclusion follows from th

ese hypotheses: Clearly label the inference rules used at every step of your proof 2. Consider the following hypotheses: H1 Vr (C()--A(x)) H2 : Vr (A() Vy B(y)) Use rules of inference to prove that the following conclusion follows C: 3r (B(x) AC()) H3: Br A(a) from these hypotheses: Clearly label the inference rules used at every step of your proof. 3. Consider the following predicate quantified formula: r y (P(x, y)P(, y)) Prove the unsatisfiability of this formula using rules of inference.

Physics
1 answer:
Igoryamba3 years ago
8 0

Answer: See attachment below

Explanation:

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1. A car’s velocity increases from 4.0 m/s to 36 m/s over a 4.0 second time interval. What is its average acceleration?
Finger [1]

Explanation:

36-4/4= 9 m/squared. meter per squared.

acceleration unit is meter per second Square.equation is velocity by time.for average final(36) minus initial(4)

7 0
3 years ago
During energy conversions, some energy is always lost as _____. light , heat , electricity or chemical energy?
masya89 [10]
Heat because in a machine heat must come out to follow the law of conservation of energy
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3 years ago
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What would be Kelley's weight be in newtons if her mass was 70 kilograms
ad-work [718]

Answer:

Kelly's weight would be 688.47 Newtons.

Explanation:

1 Kilogram would be 9.81 Newtons.

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3 years ago
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A cat leaps into the air to catch a bird with an initial speed of 2.74 m/s at an angle of 60.0° above the ground. What is the hi
Volgvan

Answer: D. 0.29 m

Explanation:

We will use the following equations to describe the leap of the cat:

y=V_{o}sin\theta t-\frac{gt^{2}}{2}   (1)

V_{y}=V_{oy}-gt   (2)

Where:

y  is the height of the cat  

V_{oy}=V_{o}sin\theta is the cat's initial velocity

\theta=60\°

g=9.8m/s^{2}  is the acceleration due gravity

t is the time

V_{y} is the y-component of the velocity

Now the cat will have its maximum height y_{max} when V_{y}=0. So equation (2) is rewritten as:

0=V_{oy}-gt   (3)

Finding t:

t=\frac{V_{oy}}{g}=\frac{V_{o}sin\theta}{g}   (4)

t=\frac{2.74 m/s sin(60\°)}{9.8m/s^{2}}   (5)

t=0.24 s   (6)

Substituting (6) in (1):

y_{max}=(2.74 m/s)sin(60\°) (0.24 s)-\frac{(9.8m/s^{2})(0.24 s)^{2}}{2}   (7)

Finally:

y_{max}=0.287 m \approx 0.29 m   (8)

3 0
3 years ago
A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle
tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

-h= vsinA\times t-gt^2/2

putting values h=15 m, v=0.8

-15 = 0.8vt - 4.9t^2  ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

-15 = 0.8\times40/0.6 - 4.9t^2

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) =  17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = -H =v sinA t - gt^2/2

⇒ 4.9t^2 - 8.9\times0.8t - 100 = 0

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0
3 years ago
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