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3 years ago

A quantity found by multiplying the force by the distance moved

1 answer:

7 0

The quantity that is calculated from the product of the force and the distance traveled due to the force is called work. It has SI units of Joules (J) which is equivalent to Newton-meter (N-m). It is the energy that happens when an object is being moved by an external force.

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MA of the first class lever may be equal to, greater than 1.Why

Anika [276]

Answer:

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Explanation:

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3 years ago

PLEASE ANSWER, HURRY! THANKS :) A green object will absorb ____________________ light and reflect ____________________ light. Ex

olga55 [171]

Answer:

It will absorb yellow and reflect green

8 0

3 years ago

Plz answer the question

Vikentia [17]

Answer:

Ray A = Incidence ray

Ray B = Reflected ray

Explanation:

From the law of reflection,

Normal: This is the line that makes an angle of 90° with the reflecting surface.

Ray A is the incidence ray: This is the ray that srikes the surface of a reflecting surface. The angle formed between the normal and the incidence ray is called the incidence angle

Ray B is the reflected ray: This is the ray leaves the surface of a reflecting surface. The angle formed between the reflected ray and the normal is called reflected angle

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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

Read 2 more answers

Both segments of the wire are made of the same metal. Current I1 flows into segment 1 from the left. How does current I1 in segm

Andrei [34K]

Answer:

current I1 = current I2

Explanation:

since the wire is made up of the same material, from Kirchoff's current law sum of currents entering a particular node or segment of wire is equals to the sum of currents leaving that particular node or segment of wire

8 0

4 years ago