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3 years ago

7

Help what is the answer

1 answer:

Ainat [17]3 years ago

5 0

Answer:

C

Explanation:

$F=k\dfrac{Q_1Q_2}{r^2}= \\\\(8.99 \times 10^9)\dfrac{(-2\times 10^{-4})(8\times 10^{-4})}{0.3^2}\approx 1.6\times 10^4 N$

Hope this helps!

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Answer:

The gravity is pulling the diver downwards but the rotation of the body means gravity cant pull him down as quickly

Explanation:

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3 years ago

A pendulum is observed to complete 23 full cycles in 58 seconds. Determine the period and

Fynjy0 [20]

Answer:

Frequency=0.39Hz

Explanation:

A pendulum completes 23 full cycles in 58 seconds

A pendulum completes 1 full cycle in 58/23=2.52 seconds

Time taken(t)=2.52 seconds

Now,

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2 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

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Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

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$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

3 years ago

The gravitational field strength on earth is 10n/kg. find the weight of an object of mass 25kg

vodka [1.7K]

Answer:

250N

Explanation:

weight = Mass(in kg) × Gravitational field strength

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6 0

3 years ago

Read 2 more answers

A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20 m above the ground. Determine the

marusya05 [52]

Answer:

Explanation:

Given

Initial velocity of ball $u=10\ m/s$

height of window $h=20\ m$

Using Equation of motion

$y=ut+\frac{1}{2}at^2$

where u=initial velocity

t=time

a=acceleration

As ball is already is at a height of 20 m so

$Y=ut+\frac{1}{2}at^2+20$

$Y=10\times t+0.5\times (-9.8)t^2+20$

$Y=-4.9t^2+10t+20$

(b)highest point is obtained at v=0

$v^2-u^2=2as$

where

v=final velocity

u=initial velocity

a=acceleration

s=displacement

$(0)-10^2=2\times (-9.8)\times s$

$s=\frac{100}{19.6}$

$s=5.102\ m$

Highest Point will be $s+20=25.102\ m$

(c)Time taken when the ball hit the ground i.e. at Y=0

$-4.9t^2+10t+20=0$

$t=3.28\ s$

impact velocity $v=\sqrt{2\times 9.8\times 25.102}$

$v=22.181\ m/s$

7 0

3 years ago