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Zielflug [23.3K]
3 years ago
5

At a certain instant an object is moving to the right with speed 1.0 m/s and has a constant acceleration to the left of 1.0 m/s2

. At what later time will the object momen- tarily be at rest?
Physics
1 answer:
professor190 [17]3 years ago
5 0

Explanation:

Given that,

Initial speed of the object, u = 1 m/s

Acceleration of the object, a=-1\ m/s^2

We need to find the time when the object comes at rest (v=0). Let it is given by :

t=\dfrac{v-u}{a}

t=\dfrac{0-1\ m/s}{-1\ m/s^2}

t = 1 second

So, the object will comes to rest at 1 second. Hence, this is the required solution.

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a baby carriage is sitting at the top of a hill that is 21m high. The carriage with the baby weighs 12kg. The carriage has _____
Ket [755]
  • Height=h=21m
  • Mass=m=12kg

\\ \rm\longmapsto P.E=mgh

\\ \rm\longmapsto P.E=12(10)(21)

\\ \rm\longmapsto P.E=120(21)

\\ \rm\longmapsto P.E=2520J

8 0
3 years ago
Q|C A string with linear density 0.500 g/m is held under tension 20.0 N. As a transverse sinusoidal wave propagates on the strin
julia-pushkina [17]

A string with linear density 0.500 g/m.

Tension 20.0 N.

The maximum speed  v_{y, max}

The energy contained in a section of string 3.00 m long as a function of v_{y, max}.

We are given following data for string with linear density held under tension :

μ = 0.5 \frac{g}{m}

  = 0.5 x 10⁻³ \frac{kg}{m}

T = 20 N

If string is L = 3m long, total energy as a function of v_{y, max} is given by:

E = 1/2 x μ x L x ω² x A²

  = 1/2 x μ x L x v^{2} _{y, max}

  = 7.5 x 10⁻⁴ v^{2} _{y, max}

So, The total energy as a function of  v^{2} _{y, max} = 7.5 x 10⁻⁴ v^{2} _{y, max}

Learn more about linear density problem here:

brainly.com/question/17190616

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3 0
2 years ago
Which of the following is not one of the four muscle shapes?
zhenek [66]

Answer:

When elements bond together or when bonds of compounds are broken and form a new substance

Explanation:

3 0
3 years ago
A circular flat coil that has N turns, encloses an area A, and carries a current i has its central axis parallel to a uniform ma
bogdanovich [222]

Answer:

Torque on the coil will be ZERO

Explanation:

As we know that the magnetic moment of the closed current carrying coil is always along its axis and it is given as

M = N i A

now we know that magnetic field is also along the axis of the coil so here as we know the equation of torque given as

\tau = \vec M \times \vec B

so we have

\tau = M B sin0

\tau = 0

5 0
4 years ago
¿Juan quiere calentar una barra de aluminio de 100 g, para realizar un trabajo. La temperatura ambiente es de 27°C pero necesita
Luda [366]

Answer:

<em>709.5 cal</em>

<em></em>

Explanation:

masa m de la barra de aluminio = 100 g

temperatura ambiente = 27 ° C

<em>Asumiremos que la barra de aluminio está en equilibrio térmico con el ambiente. </em>

Esto significa que la temperatura inicial de la barra es de 27 ° C

temperatura final a la que la barra debe calentarse = 60 ° C

el aumento de temperatura ΔT será

ΔT = 60 ° C - 27 ° C = 33 ° C

capacidad calorífica específica c del aluminio = 0.215 cal/g°C

Calor C requerido = mcΔT

<em>C = 100 x 0.215 x 33 = 709.5 cal</em>

8 0
3 years ago
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