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3 years ago

What is a distraction that a pedestrian may engage in while crossing the street?

Physics

1 answer:

goldenfox [79]3 years ago

6 0

Answer:

a dog walking or their phone rings or heard a neighbor talking to them

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In 1949, dr. tony allison observed a high frequency of kenyans carrying the sickle cell allele in coastal areas and near lake vi

Lynna [10]

Dr Tony Allision hypothesize that the sickle cell disease is connected with malaria. The disease carried by the mosquito is more common in coastal area and near Lake Victoria while it is less common in the highlands. He made a further hypothesis that sickle cell disease is common in areas where malaria is also common.

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4 years ago

What other factor affected the gravitational force between the two objects in this simulation?

neonofarm [45]

Answer:

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases

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3 years ago

A force of 20N acts on a 3.5kg mass for 10s. What is the change in the speed of the object? *Hint find the impulse first*

Elina [12.6K]

Answer:

<h2>I don't know sorry bro .. .... please make me braniest </h2>

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3 years ago

A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and

iogann1982 [59]

Answer:

$p_2 = 1.76 atm$

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

$\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}$

Putting the values

$\frac{1.15*2.37}{280} =\frac{p_2 *1.68}{304}$

$9.733*10^{-3} = \frac{p_2 *1.68}{304}$

$9.733*10^{-3}*304 = p_2*1.68$

$\frac{9.733*10^{-3}*304}{1.68} =p_2$

$p_2= 1.76 atm$

7 0

4 years ago

A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

Explanation:

m = Mass of object = $\dfrac{mg}{g}$

mg = Weight of object = 20 N

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

8 0

3 years ago