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3 years ago

A pelican flying along a horizontal path drops

Physics

1 answer:

Ludmilka [50]3 years ago

8 0

Answer:

The initial speed of the pelican is 8.81 m/s.

Explanation:

Given;

height of the pelican, h = 5.0 m

horizontal distance, X = 8.9 m

The time of flight is given by;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*5}{9.81} } \\\\t = 1.01 \ s$

The initial horizontal speed of the pelican is given by;

X = vₓt

vₓ = X / t

vₓ = 8.9 / 1.01

vₓ = 8.81 m/s

Therefore, the initial speed of the pelican is 8.81 m/s.

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Dmitry_Shevchenko [17]

Weight is caused by the influence of gravity on matter/mass. Weight is given by the formulae;

Weight = Mass * Gravitational acceleration

On earth, the gravitation acceleration is a constant 10m/s2

Therefore the weight will be influenced by mass of the object in a direct relationship.

An increase in mass will cause a proportionate increase in weight.

Explanation:

Gravity is not constant across the universe with larger celestial bodies have a greater gravitational pull. Therefore an object with the same mass on earth may weigh differently on another celestial body with different gravitational acceleration. An example is the Moon. Due to the fact that the Moon has approximately ¹/3 of earth’s gravitational force, astronauts weight ¹/3 times less on the moon than on earth.

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4 years ago

An object moves with constant acceleration 3.10 m/s2 and over a time interval reaches a final velocity of 12.4 m/s. If its initi

Harrizon [31]

Answer:

Explanation:

Given:

a = 3.10 m/s^2

vf = 12.4 m/s

vi = -6.2 m/s

t = (vf - vi)/a

= (12.4 + 6.2)/3.1

= 6 s

displacement = (vf - vi)*t

= (12.4 + 6.2) * 6

= 111.6 m.

3 0

3 years ago

Read 2 more answers

A bullet with a mass ????b=13.5 g is fired into a block of wood at velocity ????b=249. The block is attached to a spring that ha

larisa86 [58]

Answer:

$M = 0.436 kg$

Explanation:

As per energy conservation we can say that energy stored in the spring at the position of maximum compression must be equal to the kinetic energy of bullet and block system

so here we have

$\frac{1}{2}(m + M)v^2 = \frac{1}{2} kx^2$

here we know that

k = 205 N/m

x = 35 cm

$\frac{1}{2}(m + M)v^2 = \frac{1}{2}(205)(0.35)^2$

now by momentum conservation we know that

$mv_o = (m + M)v$

$v = \frac{m}{m + M} v_o$

now plug in all values in it

$v = \frac{0.0135}{0.0135 + M}(249)$

now from above equation

$\frac{1}{2}(0.0135 + M)( \frac{0.0135}{0.0135 + M}(249))^2 = 12.56$

$\frac{5.65}{0.0135 + M} = 12.56$

by solving above equation we have

$M = 0.436 kg$

3 0

4 years ago

16. A 95kg fullback, running at 8.2m/s, collided in midair with a 128 kg defensive tackle moving in the opposite direction. Both

Daniel [21]

a) 779 kg m/s

The momentum of an object is given by:

p = mv

where

m is the mass of the object

v is its velocity

For the fullback before the collision,

m = 95 kg

v = 8.2 m/s

Therefore, his momentum was:

$p=mv=(95)(8.2)=779 kg m/s$

b) -779 kg m/s

After the collision, both the fullback and the tackle come to a stop: this means that their momentum after the collision is zero,

p' = 0

The initial momentum of the fullback was

p = 779 kg m/s

Therefore, his change in momentum is

$\Delta p = p' -p =0-779 = -779 kg m/s$

where the negative sign indicates that the direction is opposite to the initial direction of motion.

c) -779 kg m/s

Here we can apply the law of conservation of momentum. In fact, the total momentum before and after the collision must be conserved. So we can write:

$p_f + p_t = p'$

where

$p_f$ is the initial momentum of the fullback

$p_t$ is the initial momentum of the tackle

p' is the final combined momentum after the collision

We already know that

$p_f = 779 kg m/s\\p' = 0$

Therefore, we can find the tackle's original momentum:

$p_t = p'-p_f = 0-(779) = -779 kg m/s$

where the negative sign indicates that the direction is opposite to the initial direction of motion of the fullback.

e) -6.1 m/s

To find the velocity of the tackle, we can use again the equation of the momentum:

p = mv

where here we have

$p=-779 kg m/s$ is the original momentum of the tackle

m = 128 kg is his mass

Solving the equation for v, we find the tackle's original velocity:

$v=\frac{p}{m}=\frac{-779}{128}=-6.1 m/s$

So, he was moving at 6.1 m/s in the direction opposite to the fullback.

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4 years ago

Monochromatic light of wavelength 720 nm is incident on parallel slits, of width 0.420 mm, separated by 0.560 mm. A diffraction

ki77a [65]

Answer:

1.81×10^-4Io

Explanation:

Check attachment

4 0

4 years ago