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3 years ago

What is Resistance?

Physics

2 answers:

FrozenT [24]3 years ago

8 0

Answer:

The electrical resistance of an object is a measure of its opposition to the flow of electric current.

Ilia_Sergeevich [38]3 years ago

3 0

The ability not to be affected by something, especially adversel

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Question 1 of 10

Marianna [84]

Answer:

Option D. ²²²₉₀Th

Explanation:

Let the unknown be ⁿₘZ. Thus, the equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

Next, we shall determine n, m and Z. This can be obtained as follow:

For n:

226 = 4 + n

Collect like terms

226 – 4 = n

222 = n

n = 222

For m:

92 = 2 + m

Collect like terms

92 – 2 = m

90 = m

m = 90

For Z:

ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th

Therefore, the complete equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th

Thus, the unknown is ²²²₉₀Th

6 0

3 years ago

the law of reflection states that the angle of incidence is equal to the angle of reflection. true or false.

mina [271]

Yes it is absolutely true!

7 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

Read 2 more answers

if a 60 kg person was standing on a platform at the surface of saturn and they jumped, they would have to push with a force grea

lidiya [134]

Answer:

A 60 kg person standing on a platform at the surface of Saturn and they jumped, they would have to push with a force greater than 540 N

Explanation:

The gravitational attraction between an object on the surface of a planet and the planet is given by the weight of the object

Therefore the force needed to be applied for an object to lift off the surface of a planet = The weight of the object

The weight of the object on the surface of a planet = m × g

Where;

m = The mass of the object

g = The strength of gravity on the planet's surface in N/kg

The given parameters are;

The mass of the person standing on a platform at the surface of Saturn, m = 60 kg

The strength of gravity on the surface of Saturn = 9 N/kg

Therefore, we have;

The weight of the person = The force greater than which the person would have to push on the surface of Saturn so as to Jump = The weight of the person on the surface of Saturn = 60 kg × 9 N/kg = 540 N

Therefore, for a 60 kg person standing on a platform at the surface of Saturn and they jumped, they would have to push with a force greater than 540 N.

4 0

3 years ago

In order to generate electricity from mechanical energy, one essential element needed to transform the energy into electrical en

Anettt [7]

Id say your answer is gonna be d

6 0

3 years ago

Read 2 more answers

What is Resistance? ​

What is Resistance?