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2 years ago

Juan lives 100m away from bill whats juan`s average speed if he reaches bill home in 50 s

1 answer:

6 0

Distance = Speed x Time, we are going to derived our equation.
You can compute for the average speed using the equation: speed = distance ÷ time.
So plugging in our data, it will look like:
Speed = distance / time
= 100 m / 50 s
= 2 m/sec
Juan average speed is 2 m / sec.

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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

2 years ago

Q¹=0,07Mc Q²=2C r=1,08 cm F=.......?

Nostrana [21]

The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = (k * q1 * q2) / (r^2) where:
q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9 N. m2/ C2
r is the distance between the two charges.

Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9 x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n

5 0

3 years ago

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about

nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s$

a) The vertical speed when the player leaves the ground is 4.45 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s$

Time taken to reach the maximum height is 0.45 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s$

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

6 0

3 years ago

Discuss impact of Newton's second law on roads

ruslelena [56]

Answer:

The second law: When a force is applied to a car, the change in motion is proportional to the force divided by the mass of the car. This law is expressed by the famous equation F = ma, where F is a force, m is the mass of the car, and a is the acceleration, or change in motion, of the car

4 0

2 years ago

What are four types of pathogens and what type of disease or sickness do they cause?

Zanzabum

Bacteria( cholera)

Virus( AIDS)

Fungi(athlete's foot)

Protist( malaria)

Pleasssssssse mark me as brainliest

7 0

2 years ago