The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:

The field at twice the distance is B/4.

8 0

3 years ago

A white billiard ball with mass mw = 1.37 kg is moving directly to the right with a speed of v = 3.14 m/s and collides elastical

Ira Lisetskai [31]

That latest value for the Angle is in Grads, not in Kilograms.

Apply law of conservation of momentum along vertical direction.

$m_1v_1sin\theta_1-m_2v_2sin\theta_2=0$

$v_2=\frac{v_1sin\theta_1}{sin\theta_2}=\frac{sin54}{sin36}v_1 = 1.376v_1$

Apply law of conservation of momentum along the horizontal direction

$m_1u_1=m_1v_1cos\theta_1+m_2v_2cos\theta_2$

$u_1=v_1(cos\theta_1+1.376cos\theta_2)$

$u_1=v_1(cos(54)+1.376cos(36))$

$u_1=(1.7) v_1$

$v_1= \frac{3.14}{1.7}=1.84m/s$

The second ball velocity is $v_2 = (1.376)(1.84)=2.531m/s$

The magnitud of final total momentum is

$m(v_1+v_2)=(1.37)(2.531+1.84)=5.98kgm/s$

The magnitude of final energy is

$\frac{1}{2}m(v^2_1+v^2_2)=\frac{1}{2}(1.37)(2.531^2+1.84^2)=6.07J$

6 0

3 years ago

Estimate the average force that a baseball pitcher's hand exerts on a 0.145-kg baseball as he throws a 40-m/s pitch. Assume the

Kitty [74]

Answer:

38.67 N

Explanation:

v = Final velocity = 40 m/s

u = Initial velocity

m = Mass of ball = 0.145kg

s = Displacement of ball = 3 m

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{40^2-0^2}{2\times 3}=\frac{800}{3}\ m/s^2$

F=ma

$\\\Rightarrow F=0.145\times \frac{800}{3}\\\Rightarrow F=38.67\ N$

∴ Average force that a baseball pitcher's hand exerts on the ball is 38.67 N

6 0

3 years ago

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