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Oxana [17]
3 years ago
6

Albert observes that Emmy's watch is ticking eanwhile, Emmy observes Albert's watch and compares it with her watch. Emmy observe

s that Albert's watch is ticking more slowly than more quickly than at the same rate as her own watch.
Albert observes that Emmy's watch is ticking_______his own watch.
a. more slowly than
b. more quickly than
c. at the same rate as
Meanwhile, Emmy observes Albert's watch and compares it with her watch.
Emmy observes that Albert's watch is ticking_______ her own watch.
a. more slowly than
b. more quickly than
c. at the same rate as
Physics
1 answer:
alukav5142 [94]3 years ago
5 0

Answer:

1) b. more quickly than

Explanation:

It is stated in the question that Albert's watch is ticking more slowly than more quickly than at the same rate as her own watch.

Answer:

2) a. more slowly than

Explanation:

Albert's watch is ticking more slowly than more quickly than at the same rate as her own watch.

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How much time will it take to do 33j of work with 11 w of power
Yuri [45]

Time required : 3 s

<h3>Further explanation </h3>

Power is the work done/second.

\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s

 

To do 33 J of work with 11 W of power

P = 11 W

W = 33 J

\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}

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What is the potential energy of a 30 Newton ball that is on the ground
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The height of a helicopter above the ground is given by h = 3.15t3, where h is in meters and t is in seconds. At t = 2.10 s, the
alexandr1967 [171]

Answer:

The mailbag will take 2.44 seconds to reach the ground.

Explanation:

The height of a helicopter above the ground is given by:

h = 3.15\times t^3

Height of helicopter at t = 2.10 seconds

h(2.10 )=3.15\times (2.10 )^3 m=29.17 m

The helicopter releases a small mailbag from the height of 29.17 m.

The initial velocity of mailbag = u = 0 m/s

Duration in which mailbag will reach the ground = T

Acceleration due to gravity = g = 9.8 m/s^2

Using second equation of motion ;

s=ut+\frac{1}{2}gt^2

We have , s = 29.17

u = 0 m/s

t = T

29.17m=0 m/s\times T+\frac{9.8 m/s^2\times T^2}{2}

Solving for T, we gte :

T = 2.44 seconds

The mailbag will take 2.44 seconds to reach the ground.

5 0
3 years ago
A 5.30 m long pole is balanced vertically on its tip. What will be the speed (in meters/second) of the tip of the pole just befo
suter [353]

Answer:

v = 17.66 m/s

Explanation:

As we know that the lower end of the pole is fixed in the ground and it start rotating about that end

so here we can say that the gravitational potential energy of the pole will convert into rotational kinetic energy of the pole about its one end

so we have

mgL = \frac{1}{2}(\frac{mL^2}{3})\omega^2

so we have

\omega = \sqrt{\frac{6g}{L}}

now we have

\omega = \sqrt{\frac{6(9.81)}{5.30}}

\omega = 3.33 rad/s

now the speed of the other tip of the pole is given as

v = \omega L

v = (3.33)(5.30) = 17.66 m/s

8 0
4 years ago
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