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What is the work done if you push a tree with<br> 50 N of force but the tree does not move?

Soloha48 [4]

Answer:

The work done is 0.

Explanation:

The reason no work is done is because the equation W = Fs.

W = work

F= force

s= displacement

In this scenario F = 50 and s= 0

Therefore.

W = 50(0)

W = 0

4 0

3 years ago

What is the logical relationship between the following two categorical propositions? Some food is not edible. Some food is inedi

sveta [45]

The answer is A. Contrapositive

3 0

3 years ago

In the aftermath of an intense earthquake, the earth as a whole "rings" with a period of 54 minutes.

lawyer [7]

The frequency of the oscillation in hertz is calculated to be 0.00031 Hz.

The frequency of a wave is defined as the number of cycles completed per second while the period refers to the time taken to complete a cycle. The frequency is the inverse of period.

So;

Period(T) = 54 minutes or 3240 seconds

Frequency (f) = T-1 = 1/T = 1/3240 seconds = 0.00031 Hz

Learn more: brainly.com/question/14588679

6 0

2 years ago

A student has the following equipment - copper wire, nail made of iron,a battery, paperclips.

Anna11 [10]

Taking the copper wire, he has to wind it around the nail made of iron. After which, he then connect both ends of the copper wire to the battery, so an electric charge travels through the wire. This is the basic electromagnet. Since a current is now flowing through the wire, a magnetic field is produced. Placing the electromagnet near the mixture of copper and iron, the magnet should attract the pieces of iron, as iron is more magnetic compared to copper. This is done over a period of time, so that only the copper pieces are left in the mixture.

3 0

3 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

3 years ago

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