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Karo-lina-s [1.5K]
3 years ago
9

Raindrops hitting side windows of a car in motion often leave diagonal streaks why​

Physics
1 answer:
stira [4]3 years ago
8 0

Answer:

The car is moving at a speed faster than the raindrops go down your window.

Explanation:

Hope this helps, have a good night.

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A 2.37-kg rock is released from rest at a height of 22.2 m. ignore air resistance and determine (a) the kinetic energy at 22.2 m
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BRAINLIEST AND MAY DOUBLE POINTS!!!
raketka [301]

Given that:

 Energy of bulb  (Work ) = 30 J,

 Time (t) = 3 sec

The power consumption =  ?

 We know that, Power can be defined as rate of doing work

              Power (P) = Work(Energy supplied) ÷ time

                               = 30 ÷ 3

                              = 10 Watts

<em> The power consumption is 10 W.</em>


3 0
3 years ago
The hierarchy of needs is the spectrum of needs ranging from basic ________ needs to ________ needs to self-actualization.
Zanzabum
The hierarchy of need was developed by Maslow. The pyramid is a motivational theory in psychology that argues that people is seeking to meet successively higher needs. The base of the pyramid is for the physiological needs to safety, belongingness, esteem, and finaly self-actualization. 
3 0
3 years ago
A single-turn circular loop of wire of radius 50 mm lies in a plane perpendicular to a spatially uniform magnetic field. During
Alinara [238K]

Answer:

ε= 7.86 mV,  Current: Anti-clockwise

Explanation:

radius= 50 mm

dt= 0.10 s

Initial magnitude of magnetic field= B1 = 200 mT

Final magnitude of magnetic field = B2 = 300 mT

Ф= B. A= BA cosα

Ф1= B1 * A * cosα

Ф1= (200*10^{-3})* \pi  * (50*10^{-3} )^2*(1)

Ф1= 0.00157 Wb

Ф2= B2 * A * cosα

Ф1= (300*10^{-3})* \pi  * (50*10^{-3} )^2*(1)

Ф2=0.00236 Wb

dФ= Ф2 - Ф1

dФ= 0.00236 - 0.00157

dФ= 0.000786 Wb

ε= \frac{d}{dt} Ф

ε=0.001786/ 0.10

ε=0.00786 v = 7.86 mV

b)

According to lenz's law the induced emf always oppose the cause producing it.

Applied field is out of the paper so the current will flow in anti-clockwise direction to produce north pole pointing toward the paper.

6 0
3 years ago
An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _
Viefleur [7K]
Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
where 
f=5 cm is the focal length
p=10 cm is the distance of the object from the mirror
q is the distance of the image from the mirror.

Rearranging the equation, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}
so, the distance of the image from the mirror is q=10 cm.

1b) The image height is given by the magnification equation:
\frac{h_i}{h_o}=- \frac{p}{q}
where h_i is the heigth of the image and h_o=1 cm is the height of the object. By rearranging the equation and using p and q, we find
h_i=-h_o  \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

3) As before, we find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

And now we can use the magnification equation to find the image height:
\frac{h_i}{h_o}=- \frac{p}{q}
Rearranging it, we find
h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm
and the negative sign means the image is inverted.
5 0
3 years ago
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