All categories

3 years ago

Which SPF level should a person look for in sunscreen as part of planning a day at the beach? pls i need it now

Physics

1 answer:

slega [8]3 years ago

3 0

Hope you do well on the test and hope this helps!

You might be interested in

El tiempo que habria que esperar para que el dia fuera 1 hora mas largo que es hoy

myrzilka [38]

Answer: el tiempo que habria que esperar para que el dia fuera 1 hora mas largo que es hoy

Explanation:

8 0

3 years ago

Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o

IrinaVladis [17]

Answer:

The net gravitational force on the mass is $1.27\times 10^{-5}$

Explanation:

We have by Newton's law of gravity the force of attraction between masses $m_{1},m_{2}$

$F_{att}=G\frac{m_{1}m_{2}}{r^{2}}$

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

$F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N$

Force of attraction between 435 kg mass and 38 kg mass is

$F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N$

Thus the net force on mass 38.0 kg is $F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}$

8 0

3 years ago

Both segments of the wire are made of the same metal. Current I1 flows into segment 1 from the left. How does current I1 in segm

Andrei [34K]

Answer:

current I1 = current I2

Explanation:

since the wire is made up of the same material, from Kirchoff's current law sum of currents entering a particular node or segment of wire is equals to the sum of currents leaving that particular node or segment of wire

8 0

3 years ago

Design your own plan for a scientific experiment. You do not have to conduct your experiment.

olganol [36]

you could use one of your scientific experiments you did over the past few years.

You could make an solinoid.

Hope this helps! Lol

8 0

3 years ago

Sherlock Holmes examines a clue by holding his magnifying glass (with

Alborosie

Answer:

Distance: -30.0 cm; image is virtual, upright, enlarged

Explanation:

We can find the distance of the image using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where:

f = 15.0 cm is the focal length of the lens (positive for a converging lens)

p = 10.0 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving for q,

$\frac{1}{q}=\frac{1}{f}-\frac{1}{q}=\frac{1}{15.0}-\frac{1}{10.0}=-0.033\\q=\frac{1}{0.033}=-30.0 cm$

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).

The magnification can be found as

$M=-\frac{q}{p}=-\frac{-30}{10}=3$

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.

Also, the fact that the magnification is positive tells us that the image is upright.

8 0

3 years ago