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sertanlavr [38]

3 years ago

14

PLEASE HELP QUICKLY 13 POINTS +7 FOR BEST ANSWER!!! calculate the density of an object that has a mass of 3.0 g and a volume of

5.0 cm3. will the object float or sink when placed in water, which has a density of 1.0 g/cm3?

2 answers:

raketka [301]3 years ago

8 0

Density = 3/5 = 0.6g/cm^3. Since the density is less than the density of water, which is 1, the object will float.

Irina18 [472]3 years ago

8 0

Density = mass / volume.

D = 3/5 = 0.6 g/cm^3.

Anything whose density is less than the density of the fluid you drop it into will float in that fluid.
This object will float in water.

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Guys I really need help with these 2 questions , it's for my final plz help asap

mars1129 [50]

Answer:

(1) Initial speed, $u=0$

Final speed, $v=165.76m/s$

Average speed, $v_a_v_g=82.87m/s$

(2) Force of gravity, $F_g=12.8\times10^1^5N$

Explanation:

(1)

Given,

Distance, $S=300meter$

Time, $t=3.62second$

It is given that drag racer started at rest.

So Initial speed, $u=0$

Using Newton's second equation of motion,

$S=ut+\frac{1}{2}at^2\\300=0+\frac{a\times3.62^2}{2} \\a=45.79m/s^2$

Newton's first equation of motion,

$v=u+at\\=0+45.79\times3.62\\=165.76 m/s$

So, Final speed, $v=165.76m/s$

Average speed is defined as totle distance divided by totle time.

$v_a_v_g=\frac{S}{t}\\=\frac{300}{3.62} \\=82.87m/s$

So, Average speed, $v_a_v_g=82.87m/s$

(2)

Gravitation: It is the natural phenomenon in which two different bodies attract each other by virtue of their masses.

According to Newton's law of gravitation, the force of attraction between two bodies is directly proportional to the masses of the bodies and inversely proportional to square of distance between centers of mass of the bodies.

$F_g\propto\frac{m_1m_2}{r^2} \\F_g=G\frac{m_1m_2}{r^2}$ where $G$ is constant of proportionality and known as gravitation constant.

Given,

Mass of Jupiter, $m_1=1.9\times10^2^7kg$

Mass of Ganymede, $m_2=1.48\times10^2^3kg$

Distance between their centers of mass, $r=1.21\times10^1^2meter$

$F_g=G\frac{m_1m_2}{r^2}\\=\frac{6.67\times10^-^1^1\times1.9\times10^2^7\times1.48\times10^2^3}{(1.21\times10^1^2)^2} \\=12.8\times10^1^5N$

So, Force of gravity, $F_g=12.8\times10^1^5N$

7 0

3 years ago

When a surface is illuminated with electromagnetic radiation of wavelength 480 nm, the maximum kinetic energy of the emitted ele

algol [13]

Answer:

Max kinetic energy for 340 nm wavelength will be $2.238\times 10^{-19}j$

Explanation:

In first case wavelength of electromagnetic radiation $\lambda =480nm=480\times 10^{-9}m$

Plank's constant $h=6.6\times 10^{-34}J-s$

Maximum kinetic energy = 0.54 eV

Energy is given by $E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{480\times 10^{-9}}=4.125\times 10^{-19}J$

We know that energy is given

$E=K_{MAX}+\Phi$ , here $\Phi$ is work function

So $4.125\times 10^{-19}=0.54\times 10^{-19}+\Phi$

$\Phi =3.585\times 10^{-19}J$

Now wavelength of second radiation = 340 nm

So energy $E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{340\times 10^{-9}}=5.823\times 10^{-19}J$

So $K_{MAX}=5.823\times 10^{-19}-3.585\times 10^{-19}=2.238\times 10^{-19}j$

6 0

4 years ago

Using the result of the preceding problem, (a) calculate the distance between fringes for 633-nm light falling on double slits s

vekshin1

To solve this problem it is necessary to apply the concepts related to the concept of superposition and the fringe separation for double slit experiment.

The equation can be written as

$\Delta y = \frac{x\lambda}{d}$

Where

$\Delta y$ = Distance between fringes

x = distance between slits and screen

d = Distance between slits

$\lambda$ = Wavelength

Our values are given as

d= 0.08mm

x =3m

$\lambda = 633nm$

In this way replacing in the equation,

$\Delta y = \frac{x\lambda}{d}$

$\Delta y = \frac{3m(633nm*(\frac{1*10^{-9}}{1nm}))}{0.08*(\frac{1*10^{-3}m}{1mm})}$

$\Delta y = 2.37*10^{-2}m$

$\Delta y = 2.37cm$

Therefore the distance between the fringes is 2.37cm

PART B) For the case in which it is submerged in water it is necessary to apply the relationship of the fringes with the index of refraction therefore

$\Delta Y_2 = \frac{\Delta Y_1}{n}$

$\Delta Y_2 = \frac{2.37}{1.33}$

$\Delta Y_2 = 1.78cm$

3 0

4 years ago

A silver bar 0.125 meter long is subjected to a temperature change from 200°C to 100°C. What will be the length of the bar after

Delicious77 [7]

$\Delta L= \alpha L_0 (T_f-T_i)$

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248

D

5 0

3 years ago

Read 2 more answers

Where in the dive above would kinetic energy and potential energy be equal?

Diano4ka-milaya [45]

Answer: The sum of an object's potential and kinetic energies is called the object's mechanical energy. As an object falls its potential energy decreases, while its kinetic energy increases. The decrease in potential energy is exactly equal to the increase in kinetic energy.

6 0

3 years ago