All categories

2 years ago

what is the magnitude of g at a height above earth surface where free fall acceleration equal 6.5m/s2

1 answer:

8 0

The magnitude of gravity is expressed in terms of its acceleration. So the magnitude of ' g ' at that altitude is exactly 6.5 m/s^2.

You might be interested in

How are Newton's 3 laws related?

Allisa [31]

Answer:Newton's three laws of motion relate to each other in that they lay a foundation for the principles of things in motion, then build upon that foundation. For example, the first law of motion,...

Explanation: WEEEEEEEEEEEEEEEEEEEEWOOOOOOOOOOOOOOOOWWWWWWWWWWWWWWWWWWWWWWOOOOOOOOOOOOOOOOOWWWWWWWWWWWWWWWWOOOOOOOOOOOOOOWWWWWWWWWWWWWWWWWWWWWWWWWWWWO-EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

3 0

2 years ago

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

The height of the cylinder is $h = 0.588\ m$

The face Area is $A = 4.19 \ m^2$

The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

4 0

2 years ago

Will an object with the density of 0.66g float?

Firdavs [7]

Yes yes it will............

6 0

2 years ago

Last week we investigated the black hole at the center of our galaxy, with massMX= 8.5×1036kg. An object whose size is on the or

swat32

Answer:

$d=2.38*10^{13}m$

Explanation:

We know the mass of the hole, so we define as,

$M_x= {8.5*10^{36}}$ kg

For which centripetal force is equal to gravitational force

$\frac{mv^2}{r}=\frac{GMm}{d^2}$

Angular velocity is equal to v/r,

$w^2r=\frac{6.67*10^{-11}*8.5*10^{36}}{d^2}$

As r=1 and w=1, we clear to d,

$d^2=5.6695*10^{26}$ m

$d=2.38*10^{13}m$

6 0

2 years ago

The temperatureTdin degrees Fahrenheit in terms of the Celsius temperaturedis given by=Td+9/5d+32.The temperatureCvin degrees Ce

Shalnov [3]

Answer:

$T_F=\frac{9}{5}(T_K-273)+32$

Explanation:

Since the temperature in degrees Fahrenheit in terms of the Celsius is given by the formula $T_F=\frac{9}{5}T_C+32$ and the temperature in degrees Celsius in terms of the Kelvin temperature is given by the formula $T_C=T_K-273$ , we can use the second formula and substitute it straight into the first formula (since a simplification is not being asked), obtaining $T_F=\frac{9}{5}(T_K-273)+32$ .

7 0

2 years ago