I need help on this!

The gravitional energy of an object is always measured realative to the ?

zaharov [31]

I would say that the answer would be MASS.

4 0

2 years ago

Read 2 more answers

The centripetal force acting on the space shuttle

KIM [24]

The centripetal force acting on the space shuttle as it orbits Earth is equal to the shuttles momentum

4 0

3 years ago

Calculate the most probable speed of an ozone molecule in the stratosphere

Marysya12 [62]

Answer:

$v_{mp}=305.83 m/s$

Explanation:

The temperature in stratosphere is generally about 270 K

molecular weight of an ozone molecule = 48 gm/mole

now formula for most probable velocity

$v_{mp}= \sqrt{\frac{2RT}{M} }$

plugging the values we get

$v_{mp}= \sqrt{\frac{2\8.314\times270}{48} }$

$v_{mp}=305.83 m/s$

7 0

3 years ago

For a wire has a circular cross section with a radius of 1.23mm.

Mila [183]

Answer:

$5.731\times 10^{-5}\ m/s$

Decrease

Explanation:

I = Current = 3.7 A

e = Charge of electron = $1.6\times 10^{-19}\ C$

n = Conduction electron density in copper = $8.49\times 10^{28}\ electrons/m^3$

$v_d$ = Drift velocity of electrons

r = Radius = 1.23 mm

Current is given by

$I=neAv_d\\\Rightarrow v_d=\dfrac{I}{neA}\\\Rightarrow v_d=\dfrac{3.7}{8.49\times 10^{28}\times 1.6\times 10^{-19}\times \pi (1.23\times 10^{-3})^2}\\\Rightarrow v_d=5.731\times 10^{-5}\ m/s$

The drift speed of the electrons is $5.731\times 10^{-5}\ m/s$

$v_d=\dfrac{I}{neA}$

From the equation we can see the following

$v_d\propto \dfrac{1}{n}$

So, if the number of conduction electrons per atom is higher than that of copper the drift velocity will decrease.

5 0

3 years ago

A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.