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Darya [45]
3 years ago
12

Two objects that are at different temperatures are added to a container of water and then the container is closed. The temperatu

re is taken at one-minute intervals. The graph shows the change in temperature for the objects over the next 7 minutes. Which statement is correct?
A) The objects are different in mass.
B) The water has more mass than both objects.
C) The water started out lower than 40°C.
D) The water released energy into the objects.
Physics
2 answers:
Blababa [14]3 years ago
8 0

Answer: C) The water started out lower than 40°C

Explanation: I saw no answer so I decided to help out and go ahead and answer the question myself to see what the right answer was. :)

Marrrta [24]3 years ago
3 0

Answer:

c

Explanation:

i did the usa test prep

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Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer sec
GaryK [48]

Mary and her younger brother Alex decide to ride the carousel at the State Fair, Mary's and Alex's  angular speed M and tangential speed vM is mathematically given as

Mary's and Alex's  angular speed=1.43

Tangential speed mary=3.22 m/s

Tangential speed alex =2.260m/s

<h3>What is Mary's and Alex's angular speed M and tangential speed vM?</h3>

Generally, the equation for angular speed is mathematically given as

w=2\pi /T\\\\Therefore\\\\w=2\pi/3.9

w = 1.61 rev/see 3.9

Centripetal acc mary = v^2/r

Centripetal acc mary  = w^2r

Centripetal acc mary = w^2x 2m

Centripetal acc. of Alex = w²x L.u

Therefore

\frac{(ac) mary }{(ac) plex}= 1.43

Hence

tang. speed V=Wr

tang. speed of mary = 1.61x2 = 3.22 m/s

tang. speed of Alex: 1.61X1·4 =2.260m/s

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user100 [1]

See this suggested solution.

1. Let a force F' is the vector sum of the forces P and Q, then it is shown on the attached picture and marked with red color.

2. according to the condition the force F holds the object, then F should have the same length as the force F' and the opposite direction.

3. using the conditions described in 2. the answer is C.

4 0
3 years ago
magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
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