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3 years ago

6

A proton and an electron are fixed in space with a separation of 859 nm. Calculate the electric potential at the midpoint betwee

n the two particles.

1 answer:

makkiz [27]3 years ago

3 0

Answer:

The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

Explanation:

Electric potential is given as;

V = E*r

where;

E is the electric field strength, = kq/r²

V = ( kq/r²)*r

V = kq/r

k is coulomb's constant = 8.99 X 10⁹ Nm²/C²

q is the charge of the particles = 1.6 X 10⁻¹⁹ C

r is the distance between the particles = 859 nm

At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm

V = (8.99 X 10⁹ * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)

V = 3.349 X 10⁻³ Volts

Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

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A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$

where

$F_{net}$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_{net}=(60)(3.1)=186 N$

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_{net}=F_a-R$

where:

$F_{net}$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_{net}=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_{net}+R=186+60=246 N$

3 0

3 years ago

If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?

Anon25 [30]

<h2>Answer:50 $ms^{-1}$ </h2>

Explanation:

Let $v_{a}$ be the airspeed.

Let $v_{w}$ be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If $v_{1}$ and $v_{2}$ are two perpendicular vectors,the resultant vector has the magnitude $\sqrt{|v|_{1}^{2}+|v|_{2}^{2}}$

Given,

$v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}$

So,the ground speed is $\sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}$

6 0

3 years ago

Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea

Kipish [7]

Answer:

6.5 m/s

Explanation:

We are given that

Distance, s=100 m

Initial speed, u=1.4 m/s

Acceleration, $a=0.20 m/s^2$

We have to find the final velocity at the end of the 100.0 m.

We know that

$v^2-u^2=2as$

Using the formula

$v^2-(1.4)^2=2\times 0.20\times 100$

$v^2-1.96=40$

$v^2=40+1.96$

$v^2=41.96$

$v=\sqrt{41.96}$

$v=6.5 m/s$

Hence, her final velocity at the end of the 100.0 m=6.5 m/s

5 0

3 years ago

Jupiter's moon Io has a radius of 1.82 ✕ 106 m and a measured gravitational field of 1.80 m/s2. What is its mass (in kg)?

AveGali [126]

Answer:

8.94*10^22 kg

Explanation:

Given that

Mass of Lo, M = ?

Radius of Lo, r = 1.82*10^6 m

Acceleration on Lo, g = 1.80 m/s²

Gravitational constant, G = 6.67*10^-11

Using the formula

g = GM/r²

Solution is attached below

Answer is 8.94*10^22 kg

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3 years ago

A student who weighs 750 newton’s runs up the steps (which have a height of 8 meters) in 13.5 seconds. How much work did the stu

nikklg [1K]

Answer:

B

Explanation:

6 0

2 years ago