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alexandr1967 [171]
3 years ago
13

When a pair of balanced forces acts on an object, the net force that results is.

Physics
1 answer:
polet [3.4K]3 years ago
7 0
D.  Equal to zero.

Because the forces balance each other.
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White light, with frequencies ranging from 4.00 x 10^14 Hz to 7.90 x 10^14 Hz, is incident on a barium surface. Given that the w
REY [17]

Answer:

0.7515875 eV

4\times 10^{14}\leq f

Explanation:

f = Maximum frequency = 7.9\times 10^{14}\ Hz

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

W = Work function = 2.52 eV

Converting to Joules

W=2.52\times 1.6\times 10^{-19}\\\Rightarrow W=4.032\times 10^{-19}\ J

Maximum photon energy is given by

E=hf\\\Rightarrow E=6.626\times 10^{-34}\times 7.9\times 10^{14}\\\Rightarrow E=5.23454\times 10^{-19}\ J

Maximum Kinetic energy is given by

K=E-W\\\Rightarrow K=5.23454\times 10^{-19}-4.032\times 10^{-19}\\\Rightarrow K=1.20254\times 10^{-19}\ J

Converting to eV

1.20254\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}=0.7515875\ eV

The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV

W=hf\\\Rightarrow f=\frac{W}{h}\\\Rightarrow f=\frac{4.032\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=6.08512\times 10^{14}\ Hz

The range of frequencies for which no electrons are ejected is

4\times 10^{14}\leq f

5 0
2 years ago
A horizontal force of 50 N is applied to the object.
USPshnik [31]

Answer:

Mass of object (m) = 5.102 kg

Explanation:

Given:

Horizontal Force (F) = 50 N

Find:

Mass of object (m) = ?

Computation:

We know that, acceleration due to gravity (g) = 9.8 m/s²

⇒ Horizontal Force (F) = mg

⇒ 50 N = m (9.8 m/s²)

⇒ Mass of object (m) =  50 / 9.8

⇒ Mass of object (m) = 5.102 kg

Mass of object (m) is 5.1 kg (Approx)

3 0
3 years ago
Dynamo converts the
Sav [38]

Answer:

\gamma  \: dynamo \: converts \: the \: mechanical \: energy \: into \: electrical \: energy \: .

Explanation:

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6 0
3 years ago
Read 2 more answers
Why are different constellations<br> of stars seen during different<br> seasons?
slamgirl [31]
Actually, they're not.  There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around.  And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night. 

Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.

Why does the night sky change at various times of the year ?  Here's how to
think about it:

The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ?  Now they're straight in the
direction of the sun.  So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.
5 0
3 years ago
Read 2 more answers
A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
skelet666 [1.2K]

Answer:

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}

Differentiate both sides with respect to d.

\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

\frac{d^{2}A}{dt^{2}}= + ve

(a) So, the area is maximum when the side of square is 29 m

(b) so, the area is minimum when the side of square is 8.76 m

8 0
3 years ago
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