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Gnom [1K]
3 years ago
5

A brick of mass 1.15 kg is attached to a thin (massless) cable and whirled around in a circle in a vertical plane. The circle ha

s a radius of 1.44 m. The cable will break if the tension exceeds 43.0 Newtons. What is the maximum speed the brick can have at the bottom of the circle before the cable will break?
Physics
1 answer:
stira [4]3 years ago
8 0

Explanation:

It is given that,

Mass of the brick, m = 1.15 kg

Radius of the circle, r = 1.44 m

The cable will break if the tension exceeds 43.0 N

Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,

T=\dfrac{mv^2}{r}+mg

\dfrac{T}{m}-g=\dfrac{v^2}{r}

v=\sqrt{(\dfrac{T}{m}-g)r}

v=\sqrt{(\dfrac{43}{1.15}-9.8)\times 1.44}

v = 6.30 m/s

So, the maximum speed of the brick at the bottom of the circle before the cable will break is 6.3 m/s. Hence, this is the required solution.

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Answer:

\lambda =6.32\ cm

Explanation:

given,

number of cycle complete (f) = 116 cycles per minute

wavelength observed at 11 m in 1.5 m.

v = \dfrac{distance}{time}

v = \dfrac{11}{1.5}

v = 7.33 m/s

\lambda = \dfrac{v}{f}

\lambda = \dfrac{7.33}{116}

\lambda =0.0632\ m

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3 years ago
Why didn't the astronauts land on the moon math worksheet answers?
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Answer and Explanation:

Here's the answer!

4 0
3 years ago
A 25 newton force applied on an object moves it 50 meters. The angle between the force and displacement is 40.0 degrees. What is
zvonat [6]
Work done is when a force is exerted to cause a displacement in a certain object. 
the equation for work done ;
work done = force applied * displacement of the object 
when the force applied is not in the same direction as that of the displacement of the object then the effect of the force is not its whole value. The force is then applied at an angle to that of the displacement of the object, then the resultant force is the force exerted* cos of the angle between force and displacement, in this instance the angle is 40 °.
the new equation is then;
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5 0
3 years ago
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While driving, your car has an initial position of 3.2 m, an initial velocity of -8.4 m/s, and
KIM [24]

Answer:

The position of the car at t = 1.5 s is at -8.1625 meters

Explanation:

The initial position of the car is 3.2 meters

The initial velocity is -8.4 m/s

The constant acceleration is 1.1 m/s²

We need to find the final position of the car at the time t = 1.5 seconds

The displacement <em>s</em> = final position - initial position

s=ut+\frac{1}{2}at^{2}, where <em>u</em> is the initial velocity, <em>a</em> is the

constant acceleration and <em>t</em> is the time

So we can find the final velocity by using the rule:

final position - initial position = ut+\frac{1}{2}at^{2}

initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s

Substitute these values in the rule

final position - 3.2 = (-8.4)(1.5)+\frac{1}{2}(1.1)(1.5)^{2}

final position - 3.2 = -12.6 + 1.2375

final position - 3.2 = -11.3625

add 3.2 for both sides

final position = -8.1625

<em>That means the car is at 8.1625 meters in opposite direction</em>

<em>The position of the car at t = 1.5 s is at -8.1625 meters </em>

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3 years ago
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Your answer is C) <span> the potential energy of an object is always greater than its kinetic energy </span>
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