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3 years ago

Please Help!!!!! According to the graph which figure shows the average speed of the horse that won the race

Physics

1 answer:

Vikentia [17]3 years ago

5 0

Average speed will be given by

$v_{avg} = \frac{total distance}{total time}$

here from graph we know that

total distance = 300 m

total time taken = 15 s

now from above equation we know that

$v_{avg} = \frac{300 m}{15 s}$

$v_{avg} = 20 m/s$

so average speed will be 20 m/s

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It is interesting to speculate on the properties of a universe with different values for the fundamental constants.

qwelly [4]

Answer:

Part a)

$\lambda = 0.345 m$

Part b)

$\Delta x = 0.274 m$

Part c)

$r = 2.8 \times 10^{11} m$

Explanation:

Part a)

De broglie wavelength is given as

$\lambda = \frac{h}{mv}$

$\lambda = \frac{1}{(0.145)(20)}$

$\lambda = 0.345 m$

Part b)

By principle of uncertainty we know that

$\Delta x \times \Delta P = \frac{h}{4\pi}$

$\Delta x \times (0.145)(21 - 19) = \frac{1}{4\pi}$

$\Delta x = 0.274 m$

Part c)

As we know that

$\frac{kq_1q_2}{r^2} = \frac{mv^2}{r}$

also we know

$mvr = \frac{nh}{2\pi}$

$v = \frac{h}{2\pi mr}$

now we have

$\frac{ke^2}{r} = \frac{mh^2}{4\pi^2m^2 r^2}$

$r = \frac{h^2}{4\pi^2mke^2}$

$r = 2.8 \times 10^{11} m$

5 0

3 years ago

An air track car with a mass of 0.75 kg and a velocity of 8.5 m/s to the right collides elastically with a 0.65kg car moving to

Sunny_sXe [5.5K]

We can do this with the conservation of momentum. The fact it is elastic means no KE is lost so we don't have to worry about the loss due to sound energy etc.

Firstly, let's calculate the momentum of both objects using p=mv:

Object 1:
p = 0.75 x 8.5 = 6.375 kgm/s

Object 2 (we will make this one negative as it is travelling in the opposite direction):
p = 0.65 x -(7.2) = -4.68 kgm/s

Based on this we know that the momentum is going to be in the direction of object one, and will be 6.375-4.68=1.695 kgm/s

Substituting this into p=mv again:

1.695 = (0.75+0.65) x v
Note I assume here the objects stick together, it doesn't specify - it should!

1.695 = 1.4v
v=1.695/1.4 = 1.2 m/s to the right (to 2sf)

8 0

3 years ago

What answer is this? Pls help I’m freaking out

kolbaska11 [484]

Answer:

do u still need help it has been 2 weeks ??

Explanation:

8 0

3 years ago

Which of the following best summarizes the quantum model of atoms?

Lelechka [254]

I believe it is A. Electrons reside in known positions in fixed orbits around the nucleus

6 0

3 years ago

Two protons move with uniform circular motion in the presence of uniform magnetic fields. Proton one moves twice as fast as prot

lubasha [3.4K]

Answer:

r₂ = 4 r

Explanation:

For this exercise let's use Newton's second law with the magnetic force

F = q v x B

bold letters indicate vectors, the magnitude of this expression is

F = q v B sin θ

in this case we assume that the angle is 90º between the speed and the magnetic field.

If we use the rule of the right hand with the positive charge, the thumb in the direction of the speed, the fingers extended in the direction of the magnetic field, the palm points in the direction of the force, which is towards the center of the circle, therefore the force is radial and the acceleration is centripetal

a = v² / r

let's use Newton's second law

F = ma

q v B = m v² / r

r = $\frac{qB}{mv}$

Let's apply this expression to our case.

Proton 1

r = \frac{qB_1}{mv_1}

Proton 2

r₂ = $\frac{q \ B_2}{m \ v_2}$

in the exercise indicate some relationships between the two protons

* v₁ = 2 v₂

v₂ = v₁ / 2

* B₂ = 2B₁

we substitute

r₂ = $\frac{q \ 2B_1}{m \ \frac{v_1}{2} }$

r₂ = 4 $\frac{qB_1}{mv_1}$

r₂ = 4 r

7 0

3 years ago