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soldi70 [24.7K]
3 years ago
14

While being thrown, a net force of 132 N acts on a baseball (mass = 140g) for a period of 4.5 x 10^-2 sec. what is the magnitude

of the change in momentum of the ball?
Physics
1 answer:
Vera_Pavlovna [14]3 years ago
5 0

Answer:

5.94 N·s

Explanation:

F = 132 N

t = 0.045 s

Impulse = Ft = (132 N)(0.045 s) = 5.94 N·s

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what is the dot product and cross product of of two vectors if the angle is between them is 90 degree?​
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\sf{\pink{\underline{\underline{\blue{GIVEN:-}}}}}

  • The angle between the two vectors is 90° .

\sf{\pink{\underline{\underline{\blue{TO\: FIND:-}}}}}

  1. The dot product of two vectors .
  2. The cross product of two vectors .

\sf{\pink{\underline{\underline{\blue{SOLUTION:-}}}}}

⚡ Let \rm{\vec{a}} and \rm{\vec{b}} are the two vectors .

✍️ We have know that,

\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:.\:\vec{b}\:=\:ab\cos{\theta}\:}}}}

Where,

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\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\cos{90^{\degree}}\:}

  • cos 90° = <u>0</u>

\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\times{0}\:}

\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:0\:}

\rm{\red{\therefore}} [1] The dot product of two vectors is “ <u>0</u> ” .

✍️ We have know that,

\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:\times\:\vec{b}\:=\:ab\sin{\theta}\:}}}}

Where,

  • θ = 90°

\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\sin{90^{\degree}}\:}

  • sin 90° = <u>1</u>

\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\times{1}\:}

\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\:}

\rm{\red{\therefore}} [2] The cross product of two vectors is “ <u>ab</u> ” .

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A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th
ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = 9.8\ m/s^2

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

                Kinetic energy K.E = \frac{1}{2}mv^2.

From above we know that -

   Kinetic energy at the bottom of the cliff = potential energy at a height h

                 \frac{1}{2}mv^2=\ mgh

                ⇒ v^2=\ 2gh

                ⇒ h=\ \frac{v^2}{2g}

                ⇒ h=\ \frac{(24.2)^2}{2\times 9.8}

                ⇒ h=\ 29.88\ m

Hence, the height of the cliff is 29.88 m

             


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