Describe how two isotopes of nitrogen differ from two nitrogen ions???PLEASE ANSWER 20 POINTS

Read the scenario, then answer the questions. a 2 kg ball is thrown upward with a velocity of 15 m/s. what is the kinetic energy

Kaylis [27]

Both answers are 225 J

3 0

3 years ago

Read 2 more answers

During a tornado in 2008 the Peachtree Plaza Westin Hotel in downtown Atlanta suffered damage. Suppose a piece of glass dropped

Darina [25.2K]

Answer:

Time = t = 6.62 s

Explanation:

Given data:

Height = h = 215 m

Initial velocity = $v_{i}$ = 0 m/s

gravitational acceleration = g = 9.8 m/s²

Time = t = ?

According to second equation of motion

$h = v_{i}t + \frac{1}{2}gt^{2}$

As initial velocity is zero, So the first term of right hand side of above equation equal to zero.

$h = \frac{1}{2}gt^{2}$

t² = $\frac{2h}{g}$

t = $\sqrt{\frac{2h}{g} }$

t = $\sqrt{\frac{(2)(215) }{9.8} }$

t = 6.62 s

3 0

3 years ago

For a wave, the _____ the amplitude, the _____ energy the wave carries.

Rudik [331]

Answer:

<h2>HIGHER & MORE OR LARGER OR MORE </h2>

HENCE, THE ANSWER IS A. :)

Explanation:

#CARRYONLEARNING

BRAINLIEST PLEASE I REALLY NEED IT

3 0

3 years ago

Read 2 more answers

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$

$mg = \mu_s ma_{min}$

$a_{min} = \dfrac{mg }{ \mu_s m}$

$a_{min} = \dfrac{g }{ \mu_s }$

where;

$\mu_s = 0.383$ and g = 9.8 m/s²

$a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }$

$\mathbf{a_{min}= 25.59 \ m/s^2}$

3 0

2 years ago

For a maximum superelevation of 0.08 ft/ft and a degree of curve of 4o, calculate the maximum safe speed for the curve assuming

Mamont248 [21]

Answer:

Explanation:

Given that

Superelation= 0.08ft/ft

Given curve= u•

Curve junction factor= 0.13

DR= 5729.57795

R = 5729.57795/D

R = 5729.57795/4

R = 1432.4ft

c + f = V^2/gG

0.08 + 0.13 = V^2 / (32*1432.4)

V^2 = 9625.728 or V = 98 ft/sec

The designed speed for a project considered is a minimum value which means the highway design elements will meet or exceed the standards for the design speed. The maximum safe speed under normal condition is significantly greater than design speed

7 0

3 years ago