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2 years ago

NEED HELP!!!!!

Physics

1 answer:

Ipatiy [6.2K]2 years ago

7 0

Answer:

when you learn to lift weights with the proper body mechanics good form becomes a habit It's true

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To produce change in motion, a force must be a(n)

Bas_tet [7]

It'd be an unbalanced force

5 0

3 years ago

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$

If we manipulate the equation we have that:

$k=\frac{2mgh}{d^{2} }$

$k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}$

$k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}$

$k=243\ \frac{N}{m}$

5 0

3 years ago

After coming down a slope, a 60-kg skier is coasting northward on a level, snowy surface at a constant 15 m>s. Her 5.0-kg cat

Vinil7 [7]

To solve this exercise, it is necessary to apply the concepts of conservation of the moment especially in objects that experience an inelastic colposition.

They are expressed as,

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

Where,

$m_1$ = mass of the skier

$m_2$ = mass of the cat

$v_1$ = initial velocity of skier

$v_2$ = initial velocity of cat

$v_f$ = final velocity of both

Re-arrange to find V_f we have,

$V_f = \frac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V_f = \frac{(60)(15)+(5)(-3.8)}{(60+5)}$

$V_f = 13.55m/s$

Once the final velocity is found it is possible to calculate the change in kinetic energy, so

$\Delta KE = KE_i-KE_f$

$\Delta KE = \frac{1}{2}(m_1v_1^2+m_2v^2_2)-\frac{1}{2}(m_1+m_2)v_f^2$

$\Delta KE = \frac{1}{2}((60)(15)^2+(5)(-3.8)^2)-\frac{1}{2}(60+5)(13.55)^2$

$\Delta KE = 819.1J$

Therefore the amount of kinetic energy converted in to internal energy is 819J

3 0

3 years ago

Imagine two fixed charges on the x axis. Charge one is +q and is located to the left of charge two which is equal to -4q. Where

givi [52]

Answer: B)To the left of the charges.

Explanation: between the charges the electric field will not cancel but will be added since electric field lines from both charges point in the same direction. To the right of the charge the -4q will take over as it’s strength overcomes the strength of the +q charge. At this point the magnitude of +q will never reach a magnitude strong enough to cancel the -4q. To the left, it is further away from -4q and is closer to +q and electric field lines point in different direction

7 0

3 years ago

A 120-V rms voltage at 1000 Hz is applied to an inductor, a 2.00-μF capacitor and a 100-Ω resistor, all in series. If the rms va

natima [27]

Answer:

The inductance of the inductor is 35.8 mH

Explanation:

Given that,

Voltage = 120-V

Frequency = 1000 Hz

Capacitor $C= 2.00\mu F$