Which statement could be defended using data from the graph ?

Draw the Free- Body diagram of the 37 kilogram glass falling to the floor in a vacuum.

Alexus [3.1K]

Answer:

362.6 N

Explanation:

$F_{g}=mg$

$F_{g}=37\cdot9.8$

$F_{g}=362.6\ N$

Therefore, the force that the glass hits the floor with is 362.6 N

5 0

3 years ago

The mass of the Earth is 5.98 × 1024 kg . A 11 kg bowling ball initially at rest is dropped from a height of 2.63 m. The acceler

jekas [21]

Answer:

v = 7.18_m/s

Explanation:

Velocity of the earth towards the ball is = velocity of the ball moving towards earth

For object in free fall, we have

Where

v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

S = height of ball above ground

v^2 = u^2 - 2×g×(-S)

= 0 + 2×9.8×2.63 = 51.55_m^2/s^2

Velocity of the ball just before it hits the ground is

v = 7.18_m/s

3 0

3 years ago

Could someone help me with this please?

ycow [4]

Answer:

you could measure several properties of

the unknown liquid and compare them with the properties of known

substances. You might observe and measure such properties as color,

odor, texture, density, boiling point, and freezing point.

6 0

3 years ago

Read 2 more answers

The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in

chubhunter [2.5K]

Answer:

The angular separation equals $0.35^{o}$

Explanation:

We have according to Snell's law

$n_{1}sin(\theta _{1})=n_{2}sin(\theta _{2})$

$\therefore \theta _{2}=sin^{-1}(\frac{n_{1}sin(\theta _{1})}{n_{2}})$

Using this equation for both the colors separately we have

$\theta _{red}=sin^{-1}(\frac{sin(23.90^{o})}{1.62})\\\\\theta _{red}=14.48^{o}$

Similarly for violet light we have

$\theta _{violet}=sin^{-1}(\frac{sin(23.90^{o})}{1.660})\\\\\theta _{violet}=14.13^{o}$

Thus the angular separation becomes

$\Delta \theta =14.48-14.13=0.35^{o}$

6 0

3 years ago

At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor

Irina18 [472]

Answer:

The charge flows in coulombs is

$dq=1.843x10^{-5}C$

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

$I=N*\frac{dF}{Rdt}$

$\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}$

$dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}$

$N=1300$ , $\beta=0.703$ , $A=\pi*r^2=\pi*0.10^2=0.01\pi m^2$ , $R=99.4+202=301.4$ Ω

$\alpha_f=14.6$ , $\alpha_i=165.4$

Replacing :

$dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}$

$dq=1.843x10^{-5}C$

5 0

3 years ago