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3 years ago

9

A ball hangs on a string. The force of tension (Ft) in the string is 15n. The force due to gravity (Fg) pulls the ball down with

a force of 15 N. Show your work.

1 answer:

Tanya [424]3 years ago

7 0

Answer: 0N

Explanation: 15N-15N =0N

they are a balanced force,meaning they canceled each other out.

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On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total

Stella [2.4K]

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

The density of air at 1 atm pressure and 20º C is 1.2 kg/m³

the volume V = 20,000/(1.2×9.8) = 1700 m³

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3 years ago

Distinguishing between Speed and Velocity.

koban [17]

Answer:

Speed: Distance per time, 400 km/h, and a scalar quantity.

Velocity: Displacement per time, 20 m/s south, and a vector quantity.

Explanation:

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3 years ago

Which of the following describes the effect of a convex mirror on light rays?

natulia [17]

B. It reflects light rays outward.

A convex mirror looks like this ⊂. So when light rays hit the outward curves, it bends the light rays outwards, and backwards, and it would appear to be coming from its center called the focus.

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3 years ago

A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5\times 10^{-

joja [24]

Answer:

$5.25\cdot 10^{40} kg m^2/s$

Explanation:

The angular momentum of the pulsar is given by:

$L=m\omega r^2$

where

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$r = 10.0 km = 1\cdot 10^4 m$ is the radius

$\omega$ is the angular speed

Given the period of the pulsar, $T=33.5\cdot 10^{-3} s$ , the angular speed is given by

$\omega=\frac{2\pi}{T}=\frac{2 \pi}{33.5\cdot 10^{-3}s}=187.5 rad/s$

And so, the angular momentum is

$L=m\omega r^2=(2.8\cdot 10^{30}kg)(187.5 rad/s)(1\cdot 10^4 m)^2=5.25\cdot 10^{40} kg m^2/s$

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3 years ago

Wil-E-Coyote drops a bowling ball off a cliff to try to catch the Roadrunner. The cliff is

PtichkaEL [24]

Answer:

t = 5.19 s

Explanation:

We have,

Height of the cliff is 132 m

It is required to find the time taken by the ball to fall to the ground. Let t is the time taken. So, using equation of kinematics as :

$y=ut+\dfrac{1}{2}gt^2\\\\\text{since}\ u=0\\\\y=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2y}{g}}\\\\t=\sqrt{\dfrac{2\times 132}{9.8}}\\\\t=5.19\ s$

So, it will take 5.19 seconds to fall to the ground.

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3 years ago