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3 years ago

The forces acting on this object are _ and the net force is __

Physics

1 answer:

8_murik_8 [283]3 years ago

7 0

The forces acting on this object are Balanced and the net force is 0 N, the correct option is option (a).

A system is said to be balanced, when the resultant of the forces acting on the system is 0.

The resultant of the forces acting in a bi-direction is calculated as

R= $\sqrt{(F_{x}^2)+F_{y}^2)}$

The force in x direction is Fx=+5-5=0

The force in x direction is Fy=+2-2=0

Therefore resultant force on the system is 0 N.

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A photon has momentum of magnitude 8.30×10−28 kg⋅m/s . Part APart complete What is the energy of this photon? Give your answer i

d1i1m1o1n [39]

Answer with Explanation:

We are given that

Momentum of photon= $8.3\times 10^{-28} kg.m/s$

a. We have to find the energy of this photon.

Speed of photon= $c=3\times 10^8 m/s$

We know that

Momentum=p= $\frac{h}{\lambda}$

Where

h= $6.63\times 10^{-34}$ J-s=Plank's constant

$\lambda=$ Wavelength of photon

$\lambda=\frac{h}{p}$

$\lambda=\frac{6.63\times 10^{-34}}{8.3\times 10^{-28}}$

$\lambda=7.99\times 10^{-7}$ m

$E=\frac{hc}{\lambda}$

$E=\frac{6.63\times 10^{-34}\times 3\times 10^8}{7.99\times 10^{-7}}$

$E=2.49\times 10^{-19}$ J

Hence, the energy of photon= $2.49\times 10^{-19}$ J

B.Energy of photon in electron volt= $\frac{2.49\times 10^{-19}}{1.6\times 10^{-19}}=1.55 eV$

Energy of photon= $1.55eV$

C.Wavelength of photon = $\lambda=7.99\times 10^{-7}m$

6 0

4 years ago

A small sphere with mass 5.00×10−7 kg and charge +8.00 μC is released from rest a distance of 0.500 m above a large horizontal i

Damm [24]

<span>Work done on charge is W = Eqd = σ/(2ε₀) x q x d = {(8.00 x 10⁻¹²)/(2 x 8.854187 x 10⁻¹²)} x 3.00 x 10⁻⁶ x (0.650 - 0.250) = 5.42116402J. KE of sphere = 0.5mv² = 0.5 x 5.00 x 10⁻⁷v² = work done by E-field on charge during its fall = 5.42116402→ v = 4657 m/s.</span>

6 0

3 years ago

A proton is confined within an atomic nucleus of diameter 3.60 fm. part a estimate the smallest range of speeds you might find f

Cerrena [4.2K]

The answer for this problem would be:
Assuming non-relativistic momentum, then you have:
ΔxΔp = mΔxΔv = h / (4)
Δv = h / (4πmΔx)
m ~ 1.67e-27 h ~ 6.62e-34,Δx = 4e-15 -->
Δv ~ 6.62e-34 / (4π * 1.67e-27 * 4e-15) ~ 7,886,270 m/s ~ 7.89e6 m/s
That's about 1% of the speed of light, the assumption that it's non-relativistic.

3 0

3 years ago

An ocean wave has a speed of 36 m/s and a frequency of 3 Hz.<br> What is its wavelength?

dsp73

Answer:

f=4Hz

Explanation:

frequency of a wave if it has a wavelength of 3.0 m and a wave speed of 12m/s

v=fλ → f=v/λ

f=12/3

f=4Hz

6 0

3 years ago

The moment of inertia of a thin ring of mass M and radius R about its symmetry axis is ICM = MR2.

monitta

Answer:

The moment of inertia of large ring is 2MR².

(A) is correct option.

Explanation:

Given that,

Mass of ring = M

Radius of ring = R

Moment of inertia of a thin ring = MR²

Moment of inertia :

Moment of inertia is the product of the mass of the ring and square of radius of the ring.

We need to calculate the moment of inertia of large ring

Using formula of moment of inertia

$I=I_{cm}+MR^2$

Where, $I_{cm}$ = moment of inertia at center of mass

M = mass of ring

R = radius of ring

Put the value into the formula

$I=MR^2+MR^2$

$I=2MR^2$

Hence, The moment of inertia of large ring is 2MR².

6 0

3 years ago

The forces acting on this object are _______ and the net force is ________

The forces acting on this object are _ and the net force is __