Which of these is a benefit of nuclear energy?

20 Accelerated motion is represented by a<br> line on a nonlinear distance time graph.

Varvara68 [4.7K]

Curved line

Explanation:

Acceleration of motion is represented by a curved line on a non-linear distance-time graph.

The acceleration of a non-linear motion is depicted using a parabola which is a curve. This implies that the velocity is constantly changing and the distance covered by the body is also changing with equal amount of time.

A plot of this will give a parabola. This can be further established using one of the equations of motion below:

x = u + $\frac{1}{2}$ at ²

This is a quadratic function where:

x is the distance

u is the initial velocity

t is the time

a is acceleration

A quadratic function gives a curved line which is a parabola.

Learn more:

Acceleration brainly.com/question/10932946

#learnwithBrainly

3 0

3 years ago

Solve each problem. Be sure to show your work and give a final answer with units rounded to the given number of significant figu

Alex Ar [27]

Answer:

1. the voltage will be 2.35×12.5 = 29.4V

2. the resistance would be 9.0/6.2= 1.45ohms

3. in series they will add up thus 4+8+12= 24ohms

4. in parallel it will be 2.18ohms

8 0

3 years ago

Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re

leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

$F_ed_e=F_ld_l$

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

$F_l=\frac{F_ed_e}{d_l}$

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

$F_l=\frac{25*60}{33}$

F_l=45.5 lbs

7 0

4 years ago

Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and

lawyer [7]

1 kilometre=1000 metre

1 hour = 3600 second

$1\ km/hr=\frac{1000}{3600} m/s$

$1\ km/hr=\frac{5}{18} m/s$

The initial velocity of car A is 35.0 km/hr i.e

$35.0\ km/hr=35*\frac{5}{18} m/s$

= 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as $25\ km/hr^2$

$=\ 25*\frac{1000}{3600*3600} m/s^2$

$=0.00192901234 m/s^2$

The time taken by car A = 15 min.

From equation of kinematics we know that-

v= u+at [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s

=11.456111106 m/s

The acceleration of B is given as $15\ km/hr^2$

$=0.00115740740740 m/s^2$

The time taken by car B =20 min

The final velocity of B is -

v= u+at

= u-at [Here a is negative due to deceleration]

=12.5 m/s +[0.0011574074074×20×60]

=13.8888888.....

=13.9

The acceleration of C is given as $40\ km/hr^2$

$=\ 0.003086419753 m/s^2$

The time taken by car C =30 min

The final velocity of C is-

v = u+at

=8.89 m/s+[0.003086419753×30×60] m/s

=14.4455555555..m/s

=14.45 m/s

The car C is decelerating.The deceleration is given as- $60\ km/hr^2$

$=0.0046296296296m/s^2$

The time taken by car D= 45 min.

The final velocity of the car D is-

v =u+at

=30.56 -[0.00462962962962×45×60]m/s

=18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

3 0

3 years ago

In an experiment, a large number of electrons are fired at a sample of neutral hydrogen atoms and observations are made of how t

kakasveta [241]

Answer:

N = 1036 times

Explanation:

The radial probability density of the hydrogen ground state is given by:

$p(r) = \frac{4r^{2} }{a_{0} ^{3} } e^{\frac{-2r}{a_{0} } }$

$p(\frac{a_{0} }{2} ) = \frac{4(\frac{a_{0} }{2} )^{2} }{a_{0} ^{3} } e^{\frac{-2(\frac{a_{0} }{2} )}{a_{0} } }$

$p(2a_{0} ) = \frac{4(2a_{0}) ^{2} }{a_{0} ^{3} } e^{\frac{-4a_{0} }{a_{0} } }$

$N = 1300\frac{p(2a_{0}) }{p(\frac{a_{0} }{2} )}$

$N = 1300\frac{(2a_{0}) ^{2}e^{\frac{-4a_{0} }{a_{0} } } }{(\frac{a_{0} }{2} )^{2} e^{\frac{-a_{0} }{a_{0} } }}$

$N = 1300(16) e^{-3}$

N = 1035.57

N = 1036 times

6 0

3 years ago