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3 years ago

A branch falls from a tree how fast is the branch moving after drug .28 seconds

1 answer:

6 0

Acceleration=9.81m/s^2
initial velocity=0m/s
time=.28s

We have to find final velocity.

The equation we use is

Final velocity=initial velocity+acceleration x time

Vf=0m/s+(9.81m/s^2)(.28s)
Vf=2.7468m/s

We would round this to:

Vf (final velocity)=2.7m/s

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Is 250 g higher than 100 g in science

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Its the same thing. Is 250 grams more then 100 grams

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What is the system that pumps blood throughout the body

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The human heart is an organ that pumps blood throughout the body via the circulatory system, supplying oxygen and nutrients to the tissues and removing carbon dioxide and other wastes.

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3 years ago

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The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur

sp2606 [1]

Answer:

a) q = -9.23 cm, b) h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

1 / f = 0.6 / 4 = 0.15

f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

1 / q = 1 / f - 1 / p

1 / q = 1 / 6.67 - 1/24

1 / q = 0.15 - 0.04167 = 0.10833

q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

M = h ’/ h = - q / p

h’= - q / p h

h’= - (-9.23) / 24.0 0.150

h’= 0.05759 cm

h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0

3 years ago

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.4 kg gibbon

Katarina [22]

Answer:

upward force acting = 261.6 N

Explanation:

given,

mass of gibbon = 9.4 kg

arm length = 0.6 m

speed of the swing

net force must provide

$F_{branch} + F_{gravity}=F_{centripetal}$

force of gravity = - mg

$F_{branch}=F_{centripetal}-F_{gravity}$

= $\dfrac{mv^2}{r} + mg$

= $m(\dfrac{3.4^2}{0.6} +9.8)$

=9 x 29.067

= 261.6 N

upward force acting = 261.6 N

7 0

3 years ago

HELP PLSSS

Andrews [41]

Given :

A 13.3 kg box sliding across the ground decelerates at 2.42 m/s².

To Find :

The coefficient of kinetic friction.

Solution :

Frictional force applied to the box is :

$f = ma$ ....1)

Also, force of friction is given by :

$f = \mu mg$ ....2)

Equating equation 1) and 2), we get :

$\mu mg = ma\\\\\mu = \dfrac{a}{g}\\\\\mu = \dfrac{2.42}{9.8}\\\\\mu = 0.247$

Therefore, the coefficient of kinetic friction is 0.247 .

8 0

3 years ago