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2 years ago

Why doesn't a ball roll on forever after being kicked at a soccer game?

Physics

2 answers:

borishaifa [10]2 years ago

7 0

Friction of the grass and gravity holding it down

Naddika [18.5K]2 years ago

6 0

Because of gravity and friction.

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A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas

vodka [1.7K]

Answer:

$v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

$F=qE$

$F_p=F_e$

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

$m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE$

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

$\frac{m_ev_e^2}{2d}=qE$

we can equals these expressions for both proton and electron, because the forces qE are the same:

$\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

4 0

2 years ago

With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by

Stells [14]

For Newton's second law, the resultant of the forces acting on the book is equal to the product between the mass of the book and its acceleration:
$\sum F = ma$ (1)

There are only two forces acting on the book:
- its weight, directed downward: mg
- the force exerted by the hand on the book, of 20 N, directed upward

so, equation (1) becomes
$mg - F = ma$
from which we can calculate the book's acceleration, a:
$a= g - \frac{F}{m}= 9.81 m/s^2 - \frac{20 N}{3 kg}=3.14 m/s^2$

7 0

2 years ago

Read 2 more answers

3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-di

irinina [24]

Answer:

The initial velocity of the water from the tank is 5.42 m/s

Explanation:

By applying Bernoulli equation between point 1 and 2

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L$

At the point 1

P₁=0 ( Gauge pressure)

V₁= 0 m/s

Z₁=3 m

At point 2

P₂=0 ( Gauge pressure)

Z₂= 0 m/s

$h_L=1.5\ m$

Now by putting the values

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L$

$Z_1-h_L=\dfrac{V_2^2}{2g}$

$3-1.5=\dfrac{V_2^2}{2\times 9.81}$

$V_2=\sqrt{2\times 1.5\times 9.81}\ m/s$

V₂= 5.42 m/s

The initial velocity of the water from the tank is 5.42 m/s

3 0

3 years ago

Two vectors A and B are directed such that there is an angle θ between them. show answer Incorrect Answer Which of the following

Troyanec [42]

Answer:

Scalar product is between ║A║ ║ B║ and -║A║ ║ B║

Explanation:

Dot product between vec A and vec B is

A.B = ║A║ ║ B║ cos θ

Here, both ║A║ and ║B║ are positive and value of cos θ depends upon θ and lies between 1 and -1

So, Scalar product is between ║A║ ║ B║ and -║A║ ║ B║

8 0

2 years ago

A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.

Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0

2 years ago