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marin [14]
3 years ago
11

Where does every piece of matter begin?

Physics
1 answer:
Margarita [4]3 years ago
6 0
Every piece of matter begins “Out of this world”
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what is the acceleration of a softball if it has a mass of 0.50kg and hits the catcher’s glove with a force of 25 N
Kitty [74]

Answer:

mass=0.50kg

force=25N

acceleration =?

Now,

force=m×a

25=0.50×a

25÷0.50=a

50=a

acceleration =50m/s^2 answer!!!!

hope this may help you!!!!

3 0
3 years ago
Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
An inductor is connected in series to a fully charged capacitor. Which of the following statements are true? Check all that appl
Luba_88 [7]

Answer:

As the capacitor is discharging, the current is increasing

Explanation:

Lets take

C= Capacitance

L=Inductance

V=Voltage

I= Current

The total energy E given as

E=\dfrac{IL^2}{2}+\dfrac{CV^2}{2}

We know that total energy E is conserved so when electric energy 1/2 CV² decreases then magnetic energy 1/2 IL²  will increases.

It means that when charge on the capacitor decreases then the current will increase.

As the capacitor is discharging, the current is increasing

3 0
3 years ago
A race car has a centripetal acceleration of 15.625 m/s2 as it goes around a curve. If the curve is a circle with radius 40 m, w
myrzilka [38]
The centripetal acceleration is given by
a_c =  \frac{v^2}{r}
where v is the tangential speed and r the radius of the circular orbit.

For the car in this problem, a_c = 15.625 m/s^2 and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
v= \sqrt{a_c r}= \sqrt{(15.625 m/s^2)(40 m)}=25 m/s
8 0
3 years ago
A large rocket has a mass of 2.00×10⁶ kg at takeoff, and its engines produce a thrust of 3.50×10⁷ N. Find its initial accelerati
Kazeer [188]

Answer:

17.5 m/s²

1.90476 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Force

F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{3.5\times 10^7}{2\times 10^6}\\\Rightarrow a=17.5\ m/s^2

Initial acceleration of the rocket is 17.5 m/s²

v=u+at\\\Rightarrow \frac{120}{3.6}=0+17.5t\\\Rightarrow t=\frac{\frac{120}{3.6}}{17.5}=1.90476\ s

Time taken by the rocket to reach 120 km/h is 1.90476 seconds

Change in the velocity of a rocket is given by the Tsiolkovsky rocket equation

\Delta v=v_{e}\ln \frac{m_0}{m_f}

where,

m_0 = Initial mass of rocket with fuel

m_f = Final mass of rocket without fuel

v_e = Exhaust gas velocity

Hence, the change in velocity increases as the mass decreases which changes the acceleration

4 0
3 years ago
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