What is the charge of an object that has lost 25 electrons?

Darren drives to school in rush hour traffic and averages 28 mph. He returns home in mid-afternoon when there is less traffic an

Gala2k [10]

Answer:

distance between school and home is 21 miles

Explanation:

given data

in rush hour speed s1 = 28 mph

less traffic speed s2 = 42 mph

time t = 1 hr 15 min = 1.25 hr

to find out

distance d

solution

we consider here distance home to school is d and t1 time to reach at school

we get here distance equation when we go home to school that is

distance = 28 × t1 .......................1

and when we go school to home distance will be

distance = 42 × ( t - t1 )

distance = 42 × ( 1.25 - t1 ) ...................2

so from equation 1 and 2

28 × t1 = 42 × ( 1.25 - t1 )

t1 = 0.75

so

from equation 1

distance = 28 × t1

distance = 28 × 0.75

distance = 21 miles

4 0

3 years ago

The fastest possible conduction velocity of action potentials is observed in ________.

irga5000 [103]

Thick, myelinated neurons

7 0

4 years ago

ਭਾਰਤ ਵਿੱਚ ਸਭ ਤੋਂ ਵੱਧ ਵਰਖਾ ਕਿੱਥੇ<br>ਹੁੰਦੀ ਹੈ।

astra-53 [7]

Answer:

I don’t understand:(

3 0

3 years ago

A particle is moving along a circular path of 2-m radius such that its position as a function of time is given by u = (5t 2) rad

OverLord2011 [107]

Answer:

Explanation:

Given

radius of path $r=2\ m$

Velocity of Particle $\theta =5t^2 rad$

where t=time in seconds

angular velocity of particle is given by

$\omega =\frac{\mathrm{d} \theta }{\mathrm{d} t}$

$\omega =2\times 5t=10\cdot t$

And angular acceleration is given by

$\alpha =\frac{\mathrm{d} \omega }{\mathrm{d} t}$

$\alpha =10 rad/s^2$

tangential acceleration is $a_t=\alpha \times r$

$a_t=10\times 2=20\ m/s^2$

Centripetal acceleration $a_c=\omega ^2\times r$

$a_c=(10t)^2\times 2=200t^2$

net acceleration is sum of tangential and centripetal force at any time t is given by

$a_{net}=\sqrt{(a_c)^2+(a_t)^2}$

$a_{net}=\sqrt{(200t)^2+(20)^2}$

$a_{net}=20\sqrt{(10t)^2+1}\ m/s$

8 0

3 years ago

A bungee jumper of mass 75kg is attached to a bungee cord of length L=35m. She walks oﬀ a platform (with no initial speed), reac

attashe74 [19]

Answer:

1. 77.31 N/m

2. 26.2 m/s

3. increase

Explanation:

1. According to the law of energy conservation, when she jumps from the bridge to the point of maximum stretch, her potential energy would be converted to elastics energy. Her kinetic energy at both of those points are 0 as speed at those points are 0.

Let g = 9.8 m/s2. And the point where the bungee ropes are stretched to maximum be ground 0 for potential energy. We have the following energy conservation equation

$E_P = E_E$

$mgh = kx^2/2$

where m = 75 kg is the mass of the jumper, h = 72 m is the vertical height from the jumping point to the lowest point, k (N/m) is the spring constant and x = 72 - 35 = 37 m is the length that the cord is stretched

$75*9.8*72 = 37^2k/2$

$k = (75*9.8*72*2)/37^2 = 77.31 N/m$

2. At 35 m below the platform, the cord isn't stretched, so there isn't any elastics energy, only potential energy converted to kinetics energy. This time let's use the 35m point as ground 0 for potential energy

$mv^2/2 = mgH$

where H = 35m this time due to the height difference between the jumping point and the point 35m below the platform

$v^2/2 = gH$

$v = \sqrt{2gH} = \sqrt{2*9.8*35} = 26.2 m/s$

3. If she jumps from her platform with a velocity, then her starting kinetic energy is no longer 0. The energy conservation equation would then be

$E_P + E_k = E_E$

So the elastics energy would increase, which would lengthen the maximum displacement of the cord

5 0

4 years ago