All categories

2 years ago

When the mass of the bottle is 0.125 kg, the KE is______ kg m2/s2.

Physics

1 answer:

tensa zangetsu [6.8K]2 years ago

5 0

Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2

Explanation:

(1) Given mass = 0.125 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2

(2) Given mass = 0.250 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2

(3) Given mass = 0.375 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2

(4) Given mass = 0.500 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2

You might be interested in

A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =

SashulF [63]

Answer:

The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

Explanation:

Given that,

Length of the current element $dl=(0.5\times10^{-3})j$

Current in y direction = 5.40 A

Point P located at $\vec{r}=(-0.730)i+(0.390)k$

The distance is

$|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}$

$|\vec{r}|=0.827\ m$

We need to calculate the magnetic field

Using Biot-savart law

$B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Put the value into the formula

$B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}$

We need to calculate the value of $\vec{dl}\times\vec{r}$

$\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k$

$\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})$

$\vec{dl}\times\vec{r}=0.000175i+0.000365k$

Put the value into the formula of magnetic field

$B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}$

$B=1.670\times10^{-10}i+3.484\times10^{-10}k$

Hence, The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

7 0

2 years ago

An argon ion laser puts out 5.0 W of continuous power at a wavelength of 532 nm. The diameter of the laser beam is 5.5 mm. If th

stepan [7]

Answer:

Number of photons travel through pin hole= $6.4*10^{17}$

Explanation:

First we will calculate the energy of single photon using below formula:

$E=\frac{h*c}{λ}$

Where :

h is plank's constant with value $6.626*10^{-34} J.s$

c is the speed of light whch is $3*10^{8}$

λ is the wave length = 532nm

$E=\frac{6.6268*10^{-34}* 3*10^{8}}{532nm}$

E= $3.73*10^{-19}$ J

Number of photons emitted per second:

$\frac{5J/s}{1 photon/3.73*10^{-19} }$

Number of photons emitted per second= $1.34*10^{19} photons/s$

$\frac{A-hole}{A-beam}$ = $\frac{\frac{pie*d-hole^2}{4}}{\frac{pie*d-beam^2}{4} }$

Where:

A-hole is area of hole

A-beam is area of beam

d-hole is diameter of hole

d-beam is diameter if beam

$\frac{A-hole}{A-beam}$ = $\frac{d-hole^2}{d-beam^2}$

$\frac{Ahole}{A-beam}$ = $\frac{1.22^2}{5.5^2}$

$\frac{A-hole}{A-beam}$ = $\frac{144}{3025}$

Number of photons travel through pin hole= $1.34*10^{19} *\frac{144}{3025}$

Number of photons travel through pin hole= $6.4*10^{17}$

7 0

2 years ago

A 38 kg crate rests on a floor. A horizontal pulling force of 170 N is needed to start the crate

Mandarinka [93]

Answer:

0.456033049

Explanation:

$F=\mu N$ where N=mg hence

$F=\mu mg$ where m is mass of object, g is acceleration due to gravity whose value is taken as $9.81 m/s^{2}$ , $\mu$ is the coefficient of static friction and F is the applied force.