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1 year ago

what torque must be applied to this object to give it an angular acceleration of 1.30 rad/s2 if it is rotated about the x axis?

Physics

1 answer:

devlian [24]1 year ago

6 0

The torque applied to an object at an angular acceleration of 1.30rad/s² rotated about the x-axis would be 7.8Nm

<h3>What is Torque?</h3>

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction.

A vector quantity is torque. The force acting on the axis determines the direction of the torque vector.

Torque is something that everyone who has ever opened a door can relate to. A door is opened by pushing on the side of the door that is furthest. It takes a lot more pressure to push on the side closest. even if both scenarios involve the same amount of work (the larger force would be applied over a smaller distance).

Learn more about torque here:

brainly.com/question/29024338

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It took 500 N of force to push a car 4 meters. How much work was done?

djverab [1.8K]

Answer:

200 J

Explanation:

W = F S

W= 500 x 4

W = 2000

6 0

3 years ago

Ender and Shen are flying at each other during a battle in space. Ender moves with a velocity v1 of 12 m/s and Shen, with a mass

Shalnov [3]

Answer:

Ender's mass = 2.25 kg

Explanation:

using law of conservation of momentum .

since there is inelastic collision

given.

$let \ \ m_{1} = m \\m_{2} = 45 kg \\v_{1} = 12 m/s \\v_{2} = 9 m/s \\\\V_{3} =6.4 m/s\\using \ \ \\m_{1}v_{1} + m_{2}v_{2} = (m_{1} +m_{2})V\\m\times12 + 45\times9 = (m+45)\times6.4\\64m -12m = 405 -288\\52m =117\\m =\frac{117}{52}\\m =2.25kg$

6 0

3 years ago

A firework is designed so that when it is fired directly upwards, it explodes and splits into three equal-massed parts at its pe

Arturiano [62]

Answer:

the third piece landed $2\,\sqrt{2}$ meters in a direction 45 degrees from the observer's right towards the front

Explanation:

Due to conservation of momentum, the momentum of the third piece must add to the other two momenta in vector from to render zero (initial momentum of the firework on a plane parallel to ground. Therefore the momentum components of the third piece must have a component to the right equal in magnitude to the momentum of the piece that traveled to the left, and a momentum component pointing to the front of the observer of equal magnitude to that of the piece that traveled behind the observer. Then the direction of the third momentum must be at 45 degrees from the observer's right towards the observer's front.

Also, the magnitude of a momentum vector of such components, is given by the Pythagorean theorem. Recall also that the magnitudes of the momentum vectors of the first two pieces must be equal since they traveled equal distances.

$|P_3|=\sqrt{P_1^2+P_2^2} \\|P_3|=\sqrt{P^2+P^2}\\|P_3|=\sqrt{2P^2}\\|P_3|=P\sqrt{2}$

Then the distance traveled by the third piece must also keep this proportionality:

$D_3=\sqrt{2} \,\,D\\D_3=\sqrt{2} \,*\,2\,\,meters\\D_3=2\,\sqrt{2} \,\,meters$

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3 years ago

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.6 multiply 1010 m (inside t

Doss [256]

We will apply the concepts related to energy conservation to develop this problem. In this way we will consider the distances and the given speed to calculate the final speed on the path from the sun. Assuming that the values exposed when saying 'multiply' is scientific notation we have the following,

$d_1 = 4.6*10^{10}m$