All categories

2 years ago

Which of the following is NOT a fundamental S.I unit? A. Meter B. Ampere C. Kelvin D. Radian

Physics

1 answer:

Llana [10]2 years ago

7 0

Explanation:

D . Radian is not a fundamental S.I unit

You might be interested in

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

2 years ago

Which statement describes how the photoelectric effect causes an electric

alina1380 [7]

Answer:

A. Light acts as particles, causing electrons on the surface it strikes to be destroyed.

Explanation:

The photoelectric effect is a phenomenon that occurs when light shined onto a metal surface causes the ejection of electrons from that metal. It was observed that only certain frequencies of light are able to cause the ejection of electrons.

8 0

2 years ago

Read 2 more answers

Do yhu kno the answer?

Charra [1.4K]

Answer:

cjkdodhfodjshdijcjsnchisbdhdosjdhsijsjf

5 0

2 years ago

Read 2 more answers

an object is producing a sound that has a wavelength in air of 2.69m. If the speed of sound in air is 346m/s, what is the freque

Misha Larkins [42]

Answer:

129.74 Hz

Explanation:

Given:

Wave velocity ( v ) = 346 m / sec

wavelength ( λ ) = 2.69 m

We have to calculate Frequency ( f ) :

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 346 / 2.69 Hz

= > f = 34600 / 269 Hz

= > f = 129.74 Hz

Hence, frequency of sound is 129.74 Hz.

5 0

2 years ago

A rock weighing 98 newtons is pushed off the edge of a bridge 50 meters above the ground. What was the potential energy of the r

lidiya [134]

Ep = 4900 because Ep = wh

4 0

3 years ago

Read 2 more answers