All categories

3 years ago

Hiii please help i’ll give brainliest if you give a correct answer please thanks!

Physics

2 answers:

lakkis [162]3 years ago

8 0

Answer: the first one

Explanation: good luck!

Alinara [238K]3 years ago

6 0

Answer:

<h3>Friction </h3>

Explanation:

Friction is a force that resists motion between surfaces that are in contact.

<h3>l hope it helps ❣❣</h3>

You might be interested in

A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.

Ivanshal [37]

Part 1)

here we know that supply took 10 s to reach the ground

so here we will have

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}\times 9.8 \times 10^2$

$y = \frac{1}{2}\times 9.8 \times 100$

$y = 490 m$

Part 2)

Here all the supply covered horizontal distance of 650 m in 10 s interval of time

so here we can say

$speed = \frac{distance}{time}$

$v = \frac{650}{10}$

$v = 65 m/s$

4 0

3 years ago

PLEASE HELP ASAP!! <br><br> What do you think engineers could do to prevent sinkholes in Florida?

padilas [110]

I think they can use more durable materials.

8 0

2 years ago

True or False<br><br> The greater the speed of an object, the less kinetic energy it possesses.

Anettt [7]

That is true because if the object is moving at Forceful speeds than it will lose more of its kinetic energy

3 0

3 years ago

Read 2 more answers

Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis

ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × $10^{-9}$ m

wavelength λ = 700 nm = 700 × $10^{-9}$ m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ ...........1

here m is 1 for single slit and d is = $\frac{1}{900*10^3 m}$

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 380 × $10^{-9}$

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 700 × $10^{-9}$

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

3 0

3 years ago

Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a

Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

$m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg$

After that, we compute the potential energy 1.00 m above the reference point:

$U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J$

Then, we compute the average kinetic energy at the specified temperature:

$K=\frac{3}{2}\frac{R}{Na}T$

Whereas $N_A$ stands for the Avogadro's number for which we have:

$K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J$

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

8 0

3 years ago