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2 years ago

8

The idea is to get as much EMF produced from the sprinter running through it. If you were the Olympic coach on a year when there

happens to be a global energy crisis, and medals were assigned based on how much EMF (or current) were produced by the sprinters, what 3 pieces of advice (one is quite obvious; the other two involve the fixed orientation of the baton and the maintained position of the baton within the circular solenoid cross-section) would you give your sprinters in order to win?

1 answer:

never [62]2 years ago

3 0

Answer:

the greater the speed, the greater the electromotive force

* The metal pole must be parallel to the field

* you must keep the ball of the field

Explanation:

To determine the advice to the runners, let's use the Farad equation to and

fem = -N $\frac{ d \phi}{dt}$ = -N $\frac{ B A Cos \theta }{dt}$

how the runners are moving

fi = B l x

fem = -N B l v

therefore the advice we can give are:

* the greater the speed, the greater the electromotive force

* The metal pole must be parallel to the field

* you must keep the ball of the field

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A rock thrown straight up takes 4.2 s to reach its maximum height what was its initial velocity

liubo4ka [24]

Consider the upward direction of motion as positive and downward direction of motion as negative.

a = acceleration due to gravity in downward direction = - 9.8 $\frac{m }{s^{2}}$

v₀ = initial velocity of rock in upward direction = ?

v = final velocity of rock at the highest point = 0 $\frac{m }{s}$

t = time to reach the maximum height = 4.2 sec

Using the kinematics equation

v = v₀ + a t

inserting the values

0 = v₀ + (- 9.8) (4.2)

v₀ = 41.2 $\frac{m }{s}$

8 0

2 years ago

A spring has a force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is

kenny6666 [7]

Answer:

$f2/f1 = \sqrt{2}$

Explanation:

From frequency of oscillation

$f = 1/2pi *\sqrt{k/m}$

Initially with the suspended string, the above equation is correct for the relation, hence

$f1 = 1/2pi *\sqrt{k/m}$

where k is force constant and m is the mass

When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

$f2 = 1/2pi *\sqrt{2k/m}$

Employing f2/ f1, we have

$f2/f1 = \sqrt{2}$

3 0

2 years ago

A zero-order reaction has a constant rate of 2.30×10−4 M/s. If after 80.0 seconds the concentration has dropped to 1.50×10−2 M,

dolphi86 [110]

Answer:

Initial concentration of the reactant = 3.34 × 10^(-2)M

Explanation:

Rate of reaction = 2.30×10−4 M/s,

Time of reaction = 80s

Final concentration = 1.50×10−2 M

Initial concentration = Rate of reaction × Time of reaction + Final concentration

= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M

Initial concentration = 3.34 × 10^(-2)M

6 0

2 years ago

The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0

2 years ago

You are moving a wagon with a friend's help you push on the left side of the wagon with 25 of force while your friend pulls from

Pavel [41]

Answer:

10N to the left side towards you

Explanation:

The net force is the resultant force that acts on a body.

Force is a push or pull on a body.

Push to left side = 25N

Pull to the right = 15N

Net force = Push to left side - Pull to the right = 25N - 15N

Net force = 10N to the left side towards you

The net force is therefore 10N to the left side towards you

5 0

2 years ago