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2 years ago

Find the height of the coconut from the ground at the moment the arrow hit it.

Physics

1 answer:

Wewaii [24]2 years ago

7 0

Answer:

$v = u + at \\ 0 = 45 \sin(20) - 9.81 \times t \\ t = 1.56s$

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A 2 kg block slides on a rough horizontal surface with muk=0.6. It has an initial velocity of 5 m/s. Use g = 10 m/s2

Irina18 [472]

Answer:

360000

Explanation:

4 0

3 years ago

Froghopper insects have a typical mass of around 12.5 mg and can jump to a height of 42.3 cm. The takeoff velocity is achieved a

Maksim231197 [3]

Answer:

2065.005 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 42.3 - 0.2 = 42.1 cm = 0.421 m

The final velocity will be the initial velocity

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.421\\\Rightarrow a=-\frac{0.421\times -9.81}{0.002}\\\Rightarrow a=2065.005\ m/s^2$

Acceleration of the frog is 2065.005 m/s²

6 0

3 years ago

Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to

sergij07 [2.7K]

Answer:

$KE=1.2036\times 10^{-12}\ J$

Explanation:

Given:

charge on the alpha particle, $q=2e=3.2\times 10^{-19}\ C$

mass of the alpha particle, $m=6.64\times 10^{-27}\ kg$

strength of a uniform magnetic field, $B=0.5\ T$

radius of the final orbit, $r=0.5\ m$

<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

$q.v.B=m.\frac{v^2}{r}$

$m.v=q.B.r$

where:

v = velocity of the alpha particle

$v=\frac{q.B.r}{m}$

$v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}$

$v=1.2048\times 10^{7}\ m.s^{-1}$

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

<u>We firstly find the relativistic mass as:</u>

$m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m$

$m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }$

$m'=6.6533\times10^{-27}\ kg$

now kinetic energy:

$KE=m'.c-m.c$

$KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2$

$KE=1.2036\times 10^{-12}\ J$

6 0

3 years ago

What is the magnitude of the electric field strength between them, if the potential 7.95 cm from the zero volt plate (and 2.05 c

Dmitriy789 [7]

Answer:

ΔVab = Ed

ΔVab = Va-Vb = Va-V0 = Va

E = Va/ d

= 413V / 0.0795 m

= 5194.97 V/M

Explanation:

the potential difference between two uniform plates is calculated by the formula of electric field.

6 0

3 years ago

The position of a full moon is located

marissa [1.9K]

Answer:

opposite the sun. between the Earth and the sun. rising perpendicular to the sun.

Explanation:

4 0

3 years ago

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