Anita is running to the right at 5 m/s , as shown in (Figure 1) . Balls 1 and 2 are thrown toward her by friends standing on the ground. According to Anita, both balls are approaching her at 16 m/s .According to her friends, with what speed was ball 1 thrown?

According to her friends, with what speed was ball 2 thrown?

Answers

Answer 1

Answer:

Final answer:

In the context of relative velocity, Ball 1 was thrown with a speed of 21 m/s and Ball 2 was thrown with a speed of 11 m/s according to her friends.

Explanation:

The problem presented involves the concept of relative velocity. The speed of the balls relative to Anita is 16 m/s and she is running at a speed of 5 m/s.

If ball 1 is thrown in the same direction as Anita is running, then the friends on ground would see the speed of the ball as the sum of its velocity relative to Anita and the speed of Anita. So, the speed of ball 1 would be 16 m/s + 5 m/s = 21 m/s. Ball 1 was thrown with a speed of 21 m/s according to her friends.

For Ball 2, if it's thrown in the opposite direction to which Anita is running, then according to her friends the speed of Ball 2 would be 16 m/s (speed of the ball relative to Anita) - 5 m/s (Anita's speed) = 11 m/s. Therefore, Ball 2 was thrown with a speed of 11 m/s according to her friends.

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Chemical principles for why we store food to keep from spoiling

Answers

There are several factors that can brought food to damage and alterations in the taste and appearance can be done in different ways brought by chemical principles in processing methods including canning, dehydration (drying), smoking and freezing; the use of packaging; and the use of food additives such as antioxidants or other preservatives.

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.2m/s and the hatchback car starts from rest and accelerates at 5m/s/s, how far away do the cars meet up again?

Answers

Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. $s_(o)$ = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.

Truck:

$v_(i)$ = 33.2 m/s, a = 0 (since the velocity is constant), $s_(o)$ = 0

Using $s =s_(o)+v_(i)t+1/2at^(2)$

s = 33.2t .......... eq (1)

Hatchback:

$a=5m/s^(2)$ , $v_(i)$ = 0 m/s (since initial velocity is zero), $s_(o)$ = 0

Using $s =s_(o)+v_(i)t+1/2at^(2)$

putting in the data we will get

$s=(1/2)(5)t^(2)$

now putting 's' value from eq (1)

$2.5t^(2)-33.2t=0$

which will give,

t = 13.28 s

so both vehicles will meet up gain after 13.28 sec.

putting t = 13.28 in eq (1) will give

s = 440.896 m

So, both vehicles will meet up again at 440.896 m.

Answer:

440.9 m

Explanation:

initial speed of the pickup truck (Up) = 33.2 m/s

acceleration of the pickup truck (ap) = 0

initial speed of the hatchback = 0

acceleration of the hatchback (ah) = 5 m/s^{2}

how far away (s) do the cars meet up again?

from the equations of motion distance covered (s) = ut + $0.5at^(2)$

distance covered by the pickup = ut + $0.5at^(2)$

where

u = initial speed of the pickup = 33.2 m/s
t = time it takes
a= acceleration of the pickup = 0
the distance covered by the pickup (s) now becomes = 33.2t + $0.5.(0).t^(2)$ = 33.2t ...equation 1

distance covered by the hatchback = ut + $0.5at^(2)$

where

u = initial speed of the hatchback = 0 m/s
t = time it takes
a= acceleration of the hatchback = 5 m/s^{2}
the distance covered by the hatchback (s) now becomes = (0)t + $0.5x5t^(2)$

= $2.5t^(2)$ ......equation 2

when the cars meet, they both would have covered the same distance, therefore

distance covered by the pickup = distance covered by the hatchback
equation 1 = equation 2

33.2t = $2.5t^(2)$
33.2 = 2.5t
t = 13.28 s

now that we have the time it takes for both cars to meet, we can put the value of the time into equation 1 or equation 2 to get the distance at which they meet

from equation 1

distance (s) = 33.2t = 33.2 x 13.28= 440.9 m

from equation 2

distance (s) = $2.5t^(2)$ = $2.5x12.8^(2)$ = 440.9 m

Which satellite gave us our first photograph of Earth from space?

Answers

The correct answer is explorer 6

Final answer:

The first satellite to capture a photograph of Earth from space was Explorer 1, launched on January 31, 1958. This and subsequent space missions like the Apollo program expanded our visual understanding and perception of our planet.

Explanation:

The first satellite to give us a photograph of the Earth from space was Explorer 1, launched on January 31, 1958. Prior to Explorer 1, the Soviet Union had launched Sputnik 1 in October 1957, but Explorer 1 was the first satellite to provide us with images of Earth from space. This marked a significant moment in human history, altering our perception of the planet.

Later on, the Apollo program further improved our visual understanding of Earth, with the Apollo 17 mission capturing the 'Blue Marble' - one of the rare full disk images of the Earth in sunlight. Such images underscored our perception of Earth as a small, yet interconnected and strikingly beautiful, celestial body floating in the vastness of space.

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Which
factors will increase the speed of a sound wave in the air?

Answers

A higher temperature, stiffer materials, and less dense materials increase the speed of sound.

Answer:

Temperature

Explanation:

A 11.0 g rifle bullet is fired with a speed of 380 m/s into a ballistic pendulum with mass 10.0 kg, suspended from a cord 70.0 cm long.a) Compute the vertical height through which the pendulum rises.(cm)
b) Compute the initial kinetic energy of the bullet;(j)
c) Compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum.(j)

Answers

a. The vertical height through which the pendulum rises is equal to 0.9 cm.

b. The initial kinetic energy of the bullet is equal to 794.2 Joules.

c. The kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum is equal to 0.883 Joules.

Given the following data:

Mass of bullet = 11.0 g

Speed = 380 m/s

Mass of pendulum = 10.0 kg

Length of cord = 70.0 cm

a. To determine the vertical height through which the pendulum rises:

First of all, we would find the final velocity by applying the law of conservation of momentum:

Momentum of bullet is equal to the sum of the momentum of bullet and pendulum.

$M_bV_b = (M_b + M_p)V$

Where:

$M_b$ is the mass of bullet.

$M_p$ is the mass of pendulum.

$V_b$ is the velocity of bullet.

V is the final velocity.

Substituting the given parameters into the formula, we have;

$0.011* 380 = (0.011+10)V\n\n4.18 = 10.011V\n\nV = (4.18)/(10.011)$

Final speed, V = 0.42 m/s

Now, we would find the height by using this formula:

$Height = (v^2)/(2g) \n\nHeight = (0.42^2)/(2* 9.8) \n\nHeight = (0.1764)/(19.6)$

Height = 0.009 meters.

In centimeters:

Height = $0.009 * 100 = 0.9 \;cm$

b. To compute the initial kinetic energy of the bullet:

$K.E_i = (1)/(2) M_bV_b^2\n\nK.E_i = (1)/(2) * 0.011 * 380^2\n\nK.E_i = 0.0055* 144400\n\nK.E_i = 794.2 \; J$

Initial kinetic energy = 794.2 Joules

c. To compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum:

$K.E = (1)/(2) (M_b + M_p)V^2\n\nK.E = (1)/(2) *(0.011 + 10) * 0.42^2\n\nK.E = (1)/(2) * 10.011 * 0.1764\n\nK.E = 5.0055 * 0.1764$

Kinetic energy = 0.883 Joules.

Answer:

a) h = 0.0088 m

b) Kb = 794.2J

c) Kt = 0.88J

Explanation:

By conservation of the linear momentum:

$m_b*V_b = (m_b+m_p)*Vt$

$Vt = (m_b*V_b)/(m_b+m_p)$

$Vt=0.42m/s$

By conservation of energy from the instant after the bullet is embedded until their maximum height:

$1/2*(m_b+m_p)*Vt^2-(m_b+m_p)*g*h=0$

$h =(Vt^2)/(2*g)$

h=0.0088m

The kinetic energy of the bullet is:

$K_b=1/2*m_b*V_b^2$

$K_b=794.2J$

The kinetic energy of the pendulum+bullet:

$K_t=1/2*(m_b+m_p)*Vt^2$

$K_t=0.88J$

BAlpha Centauri has an apparent magnitude of -0.27, whereas the apparent
magnitude of Alpha Crucis is 0.77. Identify which star appears brighter when observed from
Earth. Explain your answer.

Answers

Answer:

Alpha centauri will be brighter than Alpha Crucis .

Explanation:

Apparent magnitude of a star measures how bright a star is .

This scale is reverse logarithmic ie , the brighter the star , the lower is its magnitude . A magnitude equal to 5 scale higher represents less magnitude by a factor of 1/ 100 . In this way a difference of 1 magnitude represents a brightness ratio of 2.512 . Hence a star of brightness magnitude of 7 is less bright by a factor 2.512 than that of a star magnitude of 6 .

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