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dolphi86 [110]
3 years ago
9

Recall some things you already know about projectile motion. Does a force in the vertical direction affect the horizontal compon

ent of an object’s velocity? In this situation, can you use the horizontal velocity component to find the time required to travel some horizontal distance?
Physics
1 answer:
azamat3 years ago
7 0

Answer:

vertical force cannot change the velocity on the x-axis.   t =x/v₀ₓ

Explanation:

The force is a vector magnitude, so the forces on the x-axis affect the acceleration on this axis. Consequently a vertical force cannot change the velocity on the x-axis.

     F_{y} = m g

    Fₓ = 0

The horizontal velocity in projectile motion is constant, if we neglect the air resistance, so it can be used to find the time of a horizontal displacement

       x = v₀ₓ t

       t =x/v₀ₓ

The only magnitude that is the same for both movements is the time that is a scalar

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a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an
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We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
T_1 - W_p - W_s + T_2 =0
where
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W_s = (200kg)(9.81 m/s^2)=1962 N is the weight of the scaffold
Re-arranging, we can write the equation as
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2) Equilibrium of torques:
T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using W_p = 883 N and replacing T1 with (1), we find
2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0
from which we find
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And then, substituting T2 into (1), we find
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3 years ago
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

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(b) 1.4375 eV

Explanation:

Given:

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= 433 nm

Potential difference,

= 1.43 V

Now,

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The energy of photon will be:

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or,

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(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

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or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

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