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sweet [91]
3 years ago
5

Two soccer players kick the same 1-kg ball at the same time in opposite directions. One kicks with a force of 25 N; the other ki

cks with a force of 10 N. What is the resulting acceleration of the ball, in m/s2 Answer here I​
Physics
2 answers:
Studentka2010 [4]3 years ago
7 0

Answer:

15 (These are just here to increase the word limit)

andre [41]3 years ago
5 0

Answer:

The forces offset to produce a net force of 8 N

since the ball is 1 kg

F = ma

a = f/m = 8 N/1 kg = 8 (kg m)/(s^2 kg) = 8 m/s^2

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An electron and a 0.033 0-kg bullet each have a velocity of magnitude 495 m/s, accurate to within 0.010 0%. Within what lower li
lara31 [8.8K]

Answer:

1.170*10^-3 m

3.23*10^-32 m

Explanation:

To solve this, we apply Heisenberg's uncertainty principle.

the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

If we make Δx the subject of formula, by rearranging, we have

Δx = h / 4π * m(e).Δv

on substituting the values, we have

for the electron

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 5.67*10^-31

Δx = 1.170*10^-3 m

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 0.021

Δx = 3.23*10^-32 m

therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet

7 0
3 years ago
A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons
shepuryov [24]

Answer:

v=2.556m/s

Explanation:

From the conservation of mechanical energy

K_{E1}+U_1=K_{E2}+U_2

\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2

x_2=0.08m

v_1=0 m/s

Solve to velocity v2

m*v_2^2=k*x_1^2-k*x_2^2

v^2=\frac{k}{m}*(x_1^2-x_2^2)

v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)

v=\sqrt{6.54m^2/s^2}=2.556m/s

4 0
3 years ago
A uniform disk has a moment of inertia that is (1/2)MR2. A uniform disk of mass 13 kg, thickness 0.3 m, and radius 0.2 m is loca
kicyunya [14]
The angular momentum of an object is equal to the product of its moment of inertia and angular velocity.
L = Iω
I = 1/2 MR²
I = 1/2 x 13 x (0.2)
I = 1.3

ω = 2π/t
ω = 2π/0.3
ω = 20.9

L = 1.3 x 20.9
= 27.2 kgm²/s
6 0
3 years ago
Read 2 more answers
Jin knows that the initial internal energy of a closed system is 78 J and the final internal energy is 180 J. He also knows that
lawyer [7]

As we know by the first law of thermodynamics

Q = \Delta U + W

here we know that

Q = heat given to the system

\Delta U = U_f - U_i

W = work done by the system

now here we can say

\Delta U = 180 - 78 = 102 J

W = 64 J

now we can say that heat will be given as

Q = 64 + 102 = 166 J

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy

So best describe Jin's Error is

<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>

8 0
3 years ago
I know how the voltage increasing/not enough/stays the same and how long the battery lasts with series and parallel connections.
kirill [66]

Answer:

It comes out the positive side of the battery and goes in to the negative side of the battery

Explanation:

There are already electrons in wires in a circuit before you add the battery. By adding the battery, you're giving the electrons the energy it needs to move along the circuit.

In a series circuit, the circuit is one continuous loop so there is only one path for the electrons to go - out of the positive side of the battery and around the circuit then goes back into the negative side of the battery.

However, with a parallel circuit, there are two or more ways the electrons can go so they take the path of least resistance. The electrons still go out the positive side of a battery but along the circuit, the electrons will go through the path of least resistance ( I tend to think of it like a net with holes in it - the lower the resistance the bigger the holes for the electrons to go through so more can fit in a set amount of time ) but the electrons still go out of the positive side and in through the negative

3 0
2 years ago
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