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Sever21 [200]
3 years ago
13

An object is thrown upward from the edge of a tall building as with a velocity of 10m/s. where will the object be 3s after it is

thrown? take g=10s^2
Physics
1 answer:
KonstantinChe [14]3 years ago
6 0
20 m is the answer, hope that helps
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Explanation:

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Intro to Springs and Hooke’s law
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3 years ago
a block that is 80 degrees is places next to a 20 degree block. the warmer block heats the cooler block by the process of... A)c
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6 0
3 years ago
Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a
andrey2020 [161]

Explanation:

It is given that,

Force, F=(-8\ N)i+(6\ N)j

Position vector, r=(3i+4j)\ m

(a) The torque on the particle about the origin is given by :

\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m

(b) To find the angle between r and F use dot product formula as :

F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
Read 2 more answers
A horizontal force of 400 N is exerted on a 2.0-kg ball as it rotates (at
frutty [35]

Answer:

the speed of the ball is 10 m/s

Explanation:

Given;

magnitude of exerted force, F = 400 N

mass of the ball, m = 2 kg

radius of the circle, r = 0.5

The speed of the ball is calculated by applying centripetal force formula;

F = \frac{mv^2}{r} \\\\v^2 = \frac{Fr}{m}\\\\v = \sqrt{\frac{Fr}{m}}\\\\ v = \sqrt{\frac{400*0.5}{2}}\\\\v = 10 \ m/s

Therefore, the speed of the ball is 10 m/s

6 0
3 years ago
Read 2 more answers
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