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Free_Kalibri [48]
3 years ago
11

The human body contains many examples of levers true or false

Physics
1 answer:
Vera_Pavlovna [14]3 years ago
3 0
True : <span>There are numerous third-class </span>levers<span> in the human </span>body<span>; one example can be illustrated in the elbow joint</span>
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A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. How long was the ball in the air? Rou
Slav-nsk [51]

Answer:

3.6 seconds

Explanation:

Given:

y₀ = y = 0 m

v₀ = 31 sin 35° m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 = 0 + (31 sin 35°) t + ½ (-9.8 m/s²) t²

0 = 17.78t − 4.9t²

0 = t (17.78 − 4.9t)

t = 0 or 3.63

Rounded to the nearest tenth, the ball lands after 3.6 seconds.

4 0
3 years ago
What is the importance of leaves
sdas [7]

Answer:

Leaves are  the primary source of photosynthesis. Leaves are also have a  vital function in plant processes like transpiration and guttation. I hope it helps:)

5 0
3 years ago
Consider one such cell where the magnitude of the potential difference is 65 mV, and the inner surface of the membrane is at a h
Gelneren [198K]

Answer: W = 1.04.10^{-20} J

Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.

The work to transport an ion from a lower potential side to a higher potential side is calculated by

W=q.\Delta V

q is charge;

ΔV is the potential difference;

Potassium ion has +1 charge, which means:

p = 1.6.10^{-19} C

To determine work in joules, potential has to be in Volts, so:

\Delta V=65.10^{-3}V

Then, work is

W=1.6.10^{-19}.65.10^{-3}

W=1.04.10^{-20}

To move a potassium ion from the exterior to the interior of the cell, it is required W=1.04.10^{-20}J of energy.

8 0
3 years ago
Two particles, with charges of 20.0 nC and -20.0 nC, are fixed at points with coordinates &lt;0, 4.00 cm&gt; and &lt;0, -4.00 cm
Dmitry [639]

Answer:

Explanation:

Potential energy of the system of charges

=  9 x 10⁹ x [  q₁q₂ / r₁₂ +  q₂q₃ / r₂₃ +  q₁q₃ / r₁₃ ]

here  r₁₂ ,  r₂₃ , r₁₃ are distance between 1 st and 2 nd charge , 2 nd and 3 rd charge and fist and third charge.

r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

q₁ = 20 x 10⁻⁹ C , q₂ = - 20 x 10⁻⁹ C , q₃ = 10 x 10⁻⁹ C

Potential energy  =  9 x 10⁹ x [ - 400 x 10⁻¹⁸ / .08  +  -200x10⁻¹⁸ / .04 +  200 x 10¹⁸ / .04 ]

= 9 x 10⁹  x  - 400 x 10⁻¹⁸ / .08

= 45 x 10⁻⁶ J .

b)

Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

= 9 x 10⁹ x 10 x 10⁻⁹ / .03

= 3000 V .

potential energy of fourth particle = charge x potential

= 3000 x 40 x 10⁻⁹ = 12 x 10⁻⁵ J .

kinetic energy at infinity = 12 x 10⁻⁵ J

1/2 m v² = 12 x 10⁻⁵ J

.5 x 2 x 10⁻¹³ x v² = 12 x 10⁻⁵

v² = 12 x 10⁸

v = 3.46 x 10⁴ m/s

= 9 x 10⁹

5 0
3 years ago
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.70 m/s. Her husband Bruce suddenly realiz
Katen [24]

Answer:

The velocity is v  =  6.66 \  m/s

Henrietta is at distance s=  18.17 \  m from the under the window

Explanation:

From the question we are told that

The speed of Henrietta is v=  2.70 \ m/s

The height of the window from the ground is h  =  36.5 \  m

Generally the time taken for the lunch to reach the ground assuming it fell directly under the window is

t  =  \sqrt{\frac{2 *  h }{g} }

=> t  =  \sqrt{\frac{2 *  36.5 }{9,8} }

=> t  =  2.73 \  s

Generally the time taken for the lunch to reach Henrietta is mathematically represented as

T =  t +  t_1

Here t_1 is the time duration that elapsed after Henrietta has passed below the window the value is given as 4 s

Now

T = 2.73  +  4

=> T = 6.73 \  s

Generally the distance covered by Henrietta before catching her lunch is

s=  v  *  T

=> s=  2.70  * 6.73

=> s=  18.17 \  m

Generally the speed with which Bruce threw her lunch is mathematically represented as

v  =  \frac{18.17}{2.73}

v  =  6.66 \  m/s

4 0
3 years ago
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