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3 years ago

Which has more momentum: a 30 kg bicycle moving at 10 m/s or a 300 kg car at rest?

Physics

1 answer:

Hatshy [7]3 years ago

5 0

<h2>Answer:</h2>

Momentum = mass × velocity

Momentum of bicycle = 30 × 10 = 300 kgms¯¹

Momentum of car = 300 × 0 = 0 kgms¯¹

<h3>Therefore, 300 > 0 so momentum of bicycle is more. Hence, option A is correct. </h3>

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What is the relationship between radio waves and the visible spectrum

evablogger [386]

Answer:

They both are part of electromagnetic radiation.

Radio waves have longer wavelength than visible waves.

Radio waves have lower frequency than visible waves.

Explanation:

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2 years ago

A long uniformly charged thread (linear charge density λ= 2.5 C/m) lies along the x axis in the figure.(Figure 1) A small charge

Kamila [148]

Answer 1) The electric field at distance r from the thread is radial and has magnitude

E = λ / (2 π ε° r)

The electric field from the point charge usually is observed to follow coulomb's law:

E = Q / (4 π ε° $r^{2}$ )

Now, adding the two field vectors:

$E_{thread}$ = {2.5 / (22 π ε° X 0.07 ) ; 0}

Answer 2) $E_{q}$ = {2.3 / (4 2 π ε°) ( - 7/ (√(84); -12 / (√84))

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The y-component / x-component gives the tangent of the angle with the positive x-axes.

Please refer the graph and the attachment for better understanding.

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2 years ago

A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte

vaieri [72.5K]

Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

$R = v_x T$

$R = (17.72)(6)$

$R = 106.4 m$

so the distance from the fence is given as

$d = 106.4 - 97.5$

$d = 8.86 m$

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3 years ago

If the man has initial velocity of 0.4 m/s, acceleration is 0.343 m/s^2, a time of 5.8 seconds, and experimental final velocity

agasfer [191]

Answer:

The answer is going to be C.

Explanation:

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6 0

2 years ago

A tennis player swings her 1000 g racket with a speed of 10 m/s. She hits a 60 g tennis ball that was approaching her at a speed

Mashutka [201]

(a)

consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.

m = mass of the tennis ball = 60 g = 0.060 kg

v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s

v = final velocity of the tennis ball after being hit by racket = - 39 m/s

ΔP = change in momentum of the ball

change in momentum of the ball is given as

ΔP = m (v - v₀)

inserting the above values

ΔP = (0.060) (- 39 - 20)

ΔP = - 3.54 kgm/s

hence , magnitude of change in momentum : 3.54 kgm/s

7 0

2 years ago