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Gnoma [55]
3 years ago
13

A cue ball has a mass of 0.5 kg. During a game of pool, the cue ball is struck and now has a velocity of 3 . When it strikes a s

olid ball with a mass of 0.5 kg, the cue ball comes to a complete stop. What is the new velocity of the solid ball? Round your answer to the nearest whole number.
Physics
2 answers:
photoshop1234 [79]3 years ago
8 0

Answer: 3 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum: during the collision between the two balls, the total momentum of the system before the collision and after the collision must be conserved:

p_i = p_f

The total momentum before the collision is given only by the cue ball, since the solid ball is initially at rest, therefore

p_i = m_c u_c = (0.5 kg)(3 m/s)=1.5 kg m/s

So, the final total momentum will also be

p_f = 1.5 kg m/s

And the total momentum after the collision is given only by the solid ball, since the cue ball is now at rest, therefore:

p_f = m_s v_s

from which we find the velocity of the solid ball

v_s = \frac{p_f}{m_s}=\frac{1.5 kg m/s}{0.5 kg}=3 m/s

Otrada [13]3 years ago
6 0

3, just did the assignment

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A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio
ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is v_0

Horizontal acceleration is a_x=a

At maximum height velocity is zero therefore

v_f=v_i-gt

0=v_0\sin u-gt

t=\frac{v_0\sin u}{g}

Total time of flight T=2t=\frac{2v_0\sin u}{g}

During this time horizontal range is

R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}

R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}

For maximum range \frac{\mathrm{d} R}{\mathrm{d} u}=0

\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}

\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0

\tan 2u=\frac{g}{a}

u=\frac{1}{2}tan ^{-1}\frac{g}{a}

(b)If a =10% g

a=0.1g

thus u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}

u=42.14^{\circ}

7 0
2 years ago
Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha
Inessa05 [86]

Answer:

+1.11\mu C

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

Q_1 is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to Q_2 and Q_3 at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of Q_1.

Let the electric field intensity due to Q_2 be +E_2 and that due to Q_3 be -E_3 since the charge is negative. Hence at the origin;

+E_2-E_3=0..................(1)

From equation (1) above, we obtain the following;

E_2=E_3.................(2)

From Coulomb's law the following relationship holds;

+E_2=\frac{kQ_2}{r_2^2}\\  

-E_3=\frac{kQ_3}{r_3^2}

where r_2 is the distance of Q_2 from the origin, r_3 is the distance of Q_3 from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

k can cancel out from both side of equation (3), so that we finally obtain the following;

\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)

Given;

Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

Substituting these values into equation (4); we obtain the following;

\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

6 0
3 years ago
What is a moment of a force
kirza4 [7]

Answer:

In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. The concept originated with the studies by Archimedes of the usage of levers

7 0
2 years ago
Students measured the mass of 25.0 mL of water and found it be 25.4 g. The accepted mass is 25.0 g. What is the percent error of
Andre45 [30]
Well first of all, I think the students may have been correct. 
If they didn't use distilled water, and if it wasn't exactly at 
standard temperature, then the mass of  25.0 mL  could
very well be  25.4 grams.  We don't know that there was
any 'error' in their measurement at all.
But the question says there was, so we'll do the math:

The 'error' was  (25.4 - 25.0) = +0.4 gram

As a fraction of the 'real' value, the error was

                            +0.4 / 25.0  =  +0.016 .

To change a decimal to a percent, move the
decimal point two places that way  ===> .

                           + 0.016  =  +1.6 % .

     
Their measurement was 1.6% too high.

Let's not call it an 'error'.  Let's just call it a 'discrepancy'
between the measured value and the 'accepted' value.  OK ?
4 0
3 years ago
Read 2 more answers
How to use kinetic energy to calculate velocity
BartSMP [9]

As we know the formula of kinetic energy is

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

here given that

KE = 150,000 J

mass = 120 kg

we can use this to find speed

150,000 = \frac{1}{2} * 120 * v^2

v^2 = 2500

v = 50 m/s

So speed of above object is 50 m/s

7 0
3 years ago
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