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3 years ago

A constant force of 2.5 N to the right acts on a 4.5 kg mass for 0.90 s.

Physics

1 answer:

Alborosie3 years ago

8 0

Answer:

(a) $v_f=0.5\frac{m}{s}$

(b) $v_f=-11\frac{m}{s}$

Explanation:

(a) Since a constant external force is applied to the body, it is under an uniformly accelerated motion. Using the following kinematic equation, we calculate the final velocity of the mass if it is initially at rest( $v_0=0$ ):

$v_f=v_0+at\\v_f=at(1)$

According to Newton's second law:

$F=ma\\a=\frac{F}{m}(2)$

Replacing (2) in (1):

$v_f=\frac{F}{m}t\\v_f=\frac{2.5N}{4.5kg}(0.9s)\\v_f=0.5\frac{m}{s}$

(b) In this case we have $v_0=-11.5\frac{m}{s}$ . So, we use the final velocity equation:

$v_f=v_0+at\\v_f=v_0+\frac{F}{m}t\\v_f=-11.5\frac{m}{s}+\frac{2.5N}{4.5kg}(0.9s)\\v_f=-11\frac{m}{s}$

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A bullet is fired horizontally with an initial velocity of 144.7 m/s from a tower 11 m high. if air resistance is negligible, wh

scoundrel [369]

H = 11 m, the vertical distance that the bullet falls.
Initial vertical velocity = 0
Horizontal velocity = 144.7 m/s

The time, t, taken to fall 11 m is given by
(1/2)*(9.8 m/s²)*(t s)² = (11 m)
4.9t² = 11
t = 1.4983 s

If aerodynamic resistance is ignored, the horizontal distance traveled before the bullet hits the ground is
d = (144.7 m/s)*(1.4983 s) = 216.8 m

Answer: 216.8 m

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3 years ago

What is the amplitude of an AC voltage waveform, in units of Volts, if the RMS value is 369 V?

Blizzard [7]

Answer:

521.8 V

Explanation:

The RMS value of the voltage of an AC signal is given by

$V_{rms}=\frac{V_0}{\sqrt{2}}$

where

$V_0$ is the peak voltage, which corresponds to the amplitude of the AC waveform

In this problem, we know the RMS voltage

$V_{rms}=369 V$

Therefore, we can re-arrange the previous equation to find the peak voltage (the amplitude of the waveform):

$V_0 = \sqrt{2}V_{rms}=\sqrt{2}(369 V)=521.8 V$

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3 years ago

Two 2.0kg bodies A and B collide The velocities before the collision are U1=15i+30j and U2=-10j+5.0j After the collision V1=-5.0

Svetllana [295]

Use the law of conservation of momentum. Since the momentum is a linear measure, we can treat each of the dimension separately:

i-direction:

$m_1u_{1i}+m_2u_{2i}=m_1v_{1i}+m_2v_{2i}\\v_{2i} = \frac{m_1u_{1i}+m_2u_{2i}-m_1v_{1i}}{m_2}=\frac{(2\cdot 15-2\cdot10+2\cdot5)kg\frac{m}{s}}{2kg}=10\frac{m}{s}$

j-direction:

$m_1u_{1j}+m_2u_{2j}=m_1v_{1j}+m_2v_{2j}\\v_{2j} = \frac{m_1u_{1j}+m_2u_{2j}-m_1v_{1j}}{m_2}=\frac{(2\cdot 30+2\cdot5-2\cdot20)kg\frac{m}{s}}{2kg}=15\frac{m}{s}$