When does it have kinetic energy

What is the amount of force required to accelerate a 20 kg object to 5 m/s˛?

Llana [10]

Answer:

Force = 100N

Explanation:

Force = mass × acceleration

3 0

3 years ago

A body-centered cubic lattice has a lattice constant of 4.83 Ă. A plane cutting the lattice has intercepts of 9.66 Å, 19.32 Å, a

anastassius [24]

Answer:

Miller Indices are [2, 4, 3]

Solution:

As per the question:

Lattice Constant, C = $4.83 \AA$

Intercepts along the three axes:

$\bar{x} = 9.66 \AA$

$\bar{x} = 19.32 \AA$

$\bar{x} = 14.49 \AA$

Now,

Miller Indices gives the vector representation of the atomic plane orientation in the lattice and are found by taking the reciprocal of the intercepts.

Now, for the Miller Indices along the three axes:

a = $\frac{1}{9.66}$

b = $\frac{1}{19.32}$

c = $\frac{1}{14.49}$

To find the Miller indices, we divide a, b and c by reciprocal of lattice constant 'C' respectively:

a' = $\frac{\frac{1}{9.66}}{\frac{1}{4.83}} = \frac{1}{2}$

b' = $\frac{\frac{1}{19.32}}{\frac{1}{4.83}} = \frac{1}{4}$

c' = $\frac{\frac{1}{14.49}}{\frac{1}{4.83}} = \frac{1}{3}$

7 0

3 years ago

Ron bicycles forward with an acceleration of 2.1 m/s2. If he is applying a forward force

Mila [183]

Answer:

195

Explanation:

he can only apply as much force as his body mass so it would be 195

3 0

3 years ago

The top of a tower much like the leaning bell tower at Pisa, Italy, moves toward the south at an average rate of 1.4 mm/y. The t

Gemiola [76]

Answer:

$\omega=7.16*10^{-13}\frac{rad}{s}$

Explanation:

The angular speed is given by:

$\omega=\frac{v}{r}$

Here v is the linear speed and r is the radius of the circular motion. The height of the tower is equal to the radius of the circular motion of the top of the tower, since is rotating about its base. We need to convert the given linear speed to $\frac{m}{s}$ :

$1.4\frac{mm}{y}*\frac{10^{-3}m}{1mm}*\frac{1y}{3.154*10^7s}=4.44*10^{-11}\frac{m}{s}$

Now, we calculate the angular speed:

$\omega=\frac{4.44*10^{-11}\frac{m}{s}}{62m}\\\omega=7.16*10^{-13}\frac{rad}{s}$

8 0

4 years ago

Equations to use: v= λ ∙ f v=d/t

Margarita [4]

b. 460.8 m/s

Explanation:

The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is

$f=\frac{v}{2L}$

where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:

$v=2Lf=2(0.90 m)(256 Hz)=460.8 m/s$

c. 18,000 m

Explanation:

The relationship between speed of the wave, distance travelled and time taken is

$v=\frac{d}{t}$

where

v = 6,000 m/s is the speed of the wave

d = ? is the distance travelled

t = 3 s is the time taken

Re-arranging the formula and substituting the numbers into it, we find:

$d=vt=(6,000 m/s)(3 s)=18,000 m$

3 0

3 years ago