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Phoenix [80]
3 years ago
13

What is an expression for the vertical component of this vector?

Physics
1 answer:
grandymaker [24]3 years ago
4 0
The correct choice is D .
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theoretically speaking I don't even wanna believe it's possible but if it does then then you should check for abortion

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Um corpo de massa igual a 2kg move-se com velocidade constante num plano horizontal sem atrito, conforme a figura. Em seguida, e
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An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17
Softa [21]

Answer:

distance cover is  = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

t_1 = 3.32 sec

t_2 = 5.08 sec

from equation of motion we know that

d_1 = vt_1 + \frac{1}{2} gt_1^2

where d_1 is distance covered in time t1

sod_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2=

d_1 = 110.78 m

d_2 = vt_2 + \frac{1}{2} gt_2^2

where d_2 is distance covered in time t2

d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2

d_2 = 213.31 m

distance cover is  = 213.31 - 110.78 = 102.53 m

3 0
3 years ago
When air cools, it holds _____ water than when it is warmer.
Blizzard [7]
It holds A. Less water than when it is warmer.
6 0
3 years ago
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An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elec
vaieri [72.5K]

Answer:

Speed = 575 m/s

Mechanical energy is conserved in electrostatic, magnetic and gravitational forces.

Explanation:

Given :

Potential difference, U = $-3.45 \times 10^{-3} \ V$

Mass of the alpha particle, $m_{\alpha} = 6.68 \times 10^{-27} \ kg$

Charge of the alpha particle is, $q_{\alpha} = 3.20 \times 10^{-19} \ C$

So the potential difference for the alpha particle when it is accelerated through the potential difference is

$U=\Delta Vq_{\alpha}$

And the kinetic energy gained by the alpha particle is

$K.E. =\frac{1}{2}m_{\alpha}v_{\alpha}^2 $

From the law of conservation of energy, we get

$K.E. = U$

$\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \Delta V q_{\alpha}$

$v_{\alpha} = \sqrt{\frac{2 \Delta V q_{\alpha}}{m_{\alpha}}}$

$v_{\alpha} = \sqrt{\frac{2(3.45 \times 10^{-3 })(3.2 \times 10^{-19})}{6.68 \times 10^{-27}}}$

$v_{\alpha} \approx 575 \ m/s$

The mechanical energy is conserved in the presence of the following conservative forces :

-- electrostatic forces

-- magnetic forces

-- gravitational forces

5 0
3 years ago
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