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mr_godi [17]
2 years ago
10

At t1, Car A and car B are each located at position xo moving forward at speed v. At t2, car A is located at position 2xo moving

forward at speed 3v, while car B is located at position 2xo but is moving backward at speed v. Is the average velocity of car A between t1 and t2greater than, less than, or equal to the average velocity of car B between t1 and t2?
Physics
1 answer:
wel2 years ago
3 0

Answer:

the average velocity of car A between t1 and t2greater is greater than the average velocity of B berween t1 and t2

Explanation:

Velocity is displacement over time,

Displacement is the distance covered relative to the initial starting position

For A:

at time ti, A moved from Xo to 2Xo, displacement is 2Xo.

at time t2 a moves with speed 3V, hence, his new position will be 3Xo from 2Xo which will be at 5Xo. A's displacement is 5Xo from starting point.

For B:

at time ti, B moved from Xo to 2Xo, displacement is 2Xo.

at time t2 a moves with speed V in the opposite position so he'll be back to his starting point, hence, his new position will be at Xo. A's displacement is 0 from his starting point.

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If period of the pendulum in preceding sample problem were 24s how tall would the tower be ?
frutty [35]

Answer:

So length of pendulum is 143.129 m

Explanation:

We have given period of simple pendulum is 2 sec

We have to find the length of simple pendulum

Let the length of pendulum is l

Acceleration due to gravityg=9.8m/sec^2 is

Time period is given by T=2\pi \sqrt{\frac{l}{g}}

So 24=2\times 3.14\times  \sqrt{\frac{l}{9.8}}

\sqrt{\frac{l}{9.8}}=3.821

Squaring both side

{\frac{l}{9.8}}=14.60

l =143.129 m

So length of pendulum is 143.129 m

8 0
2 years ago
Read 2 more answers
A cubical shell with edges of length a is positioned so that two adjacent sides of one face are coincident with the +x and +y ax
Bingel [31]

Answer:

Q = ba⁴ * ε₀

Explanation:

From Gauss's Law, we know that

flux Φ = Q / ε₀

where ε₀ = 8.85e-12 C²/N·m²

and also,

Φ = EAcosθ

The field is directed along the x-axis, so that all of the flux passes through the side of the cube at x = a. This means that θ = 0º, and thus

Φ = EAcos0

Φ = EA

E = bx² meanwhile, we are interested in the point where x = a, so we substitute and then

E = ba²

Since A = a² for the cube face, we have

Q / ε₀ = E * A

Q / ε₀ = ba² * a²

so that

Q = ba⁴ * ε₀

5 0
2 years ago
A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa
gregori [183]

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

\texttt{ }

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

\boxed {I = \Sigma F \times t}

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

Ek = 70\% \ Ek_o

\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )

v^2 = 70 \% \ (v_o)^2

v = - v_o \sqrt{70 \%} → <em>negative sign due to ball rebounds</em>

v = - v_o \sqrt{0.7} \texttt{ m/s}

\texttt{ }

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

\Delta p_{stone} = - \Delta p_{ball}

\Delta p_{stone} = - [ mv - mv_o ]

\Delta p_{stone} = m[ v_o - v ]

\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]

\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]

\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]

\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Average Speed of Plane : brainly.com/question/12826372
  • Impulse : brainly.com/question/12855855
  • Gravity : brainly.com/question/1724648

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0
2 years ago
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
yulyashka [42]

Answer:

178200 g mile pounds

Explanation:

Work= Force * Distance= Fh

F=ma=mg where m is mass and g is acceleration due to gravity

Work= 165 pounds *g* 1080 m=  178200 g mile pounds

5 0
2 years ago
Suppose that sunlight is incident upon both a pair of reading glasses and a pair of sunglasses. Which pair would you expect to b
Ainat [17]

Answer: the pair of sunglasses

Explanation:

A good pair of sunglasses are composed of abosorbent lenses that filter the sunlight that affects the eyes retina, especially ultraviolet (UV). So, these sunglasses are used to reduce the amount of light or radiant energy transmitted.

On the other hand, normal reading glasses (in which the lens glass has not been treated to filter ultraviolet sunlight) will let UV rays pass through.

Therefore, if both glasses are exposed to sunlight, the sunglasses are expected to be warmer by absorbing that radiant energy and preventing it from reaching the eyes.

4 0
2 years ago
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