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3 years ago

What's the distance over which a wave's shape repeats

Physics

2 answers:

gulaghasi [49]3 years ago

8 0

That's called the wave's "wavelength" .

nekit [7.7K]3 years ago

6 0

Wavelength is what it's called

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1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0

2 years ago

Rotation of the lever OA is controlled by the motion of the contacting circular disk of radius r = 300 mm whose center is given

sergiy2304 [10]

Answer:

The angular velocity is

5.64rad/s

Explanation:

This problem bothers on curvilinear motion

The angular velocity is defined as the rate of change of angular displacement it is expressed in rad/s

We know that the velocity v is given as

v= ωr

Where ω is the angular velocity

r is 300mm to meter = 0.3m

the radius of the circle

described by the level

v=1.64m/s

Making ω subject of the formula and solving we have

ω=v/r

ω=1.64/0.3

ω=5.46 rad/s

3 0

2 years ago

A 1.53-kg bucket hangs on a rope wrapped around a pulley of mass 7.07 kg and radius 66 cm. This pulley is frictionless in its ax

levacccp [35]

Answer:

$\alpha = 6.431\,\frac{rad}{s^{2}}$

Explanation:

The pulley is modelled by the Newton's Laws, whose equation of equilibrium is:

$\Sigma M = T \cdot R = \frac{1}{2}\cdot M \cdot R^{2}\cdot \alpha$

Given that tension is equal to the weight of the bucket, the angular acceleration experimented by the pulley is:

$T = \frac{1}{2}\cdot M \cdot R \cdot \alpha$

$m_{b}\cdot g = \frac{1}{2}\cdot M \cdot R \cdot \alpha$

$\alpha = \frac{2\cdot m_{b}\cdot g}{M\cdot R}$

$\alpha = \frac{2\cdot (1.53\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{(7.07\,kg)\cdot (0.66\,m)}$

$\alpha = 6.431\,\frac{rad}{s^{2}}$

7 0

2 years ago

Four +2 μC charges are placed at the positions (10 cm, 0 cm), (−10 cm, 0 cm), (0 cm, 10 cm), and (0 cm, −10 cm) such that they f

Rufina [12.5K]

Answer:

The force on the charge at the origin is 0 N .

Explanation:

All charges are positive. So, in x axis force exerted by the charge located in the position (10 cm, 0 cm) will be canceled with the force exerted by the charge located in the position (-10 cm, 0 cm). In the same way, in y axis the force exerted by the charge located in the position (0 cm, 10 cm) will be canceled with the force exerted by the charge located in the position (0 cm, -10 cm).

4 0

3 years ago

All describe the relationship of the parts of an electrical current except: _________.

kaheart [24]

C. Electrical current increases as resistance decreases

6 0

3 years ago