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3 years ago

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A 45 kg boy runs at a rate of 2.5 m/s and jumps on top of a stationary skateboard that has a mass of 4 kg. After jumping onto th

e board, the boy and the board travel off together. Assuming that momentum is conserved, what is the final speed of the boy and the board?

2 answers:

Natalija [7]3 years ago

7 0

Answer:

v = 2.29 m/s

Explanation:

As we know that the external force on the system of mass of boy + board is ZERO

So here we can use momentum conservation

now we have

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

now we have

$45 (2.5) + 4(0) = (45 + 4) v$

now we have

$v = \frac{45}{49} (2.5)$

$v = 2.29 m/s$

igomit [66]3 years ago

4 0

Answer:

V = 2.29 m/s

Explanation:

Given that,

Mass of the boy, $m_1=45\ kg$

Mass of the skateboard, $m_2=4\ kg$

Initial speed of the boy, v = 2.5 m/s

Let V is the final velocity of the boy and the board. The net momentum of the system remains constant. Using the conservation of linear momentum to find it as :

$45\times 2.5=(45+4)V$

$V=\dfrac{45\times 2.5}{(45+4)}$

V = 2.29 m/s

So, the velocity of the boat after Batman lands in it 2.29 m/s. Hence, this is the required solution.

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The wavelength of the light is 590nm

wavelength of the light=λ

Given,

The equation for a diffraction grating to use here is dsinθ = mλ

d=1/610 lines per mm=1.639* $10^{-6}$ , m=1, θ=21. $1^{0}$

λ=1.639* $10^{-6}$ * sin21. $1^{0}$ /1 =590nm

<h3>Wavelength </h3>

The wavelength, which in physics refers to the length over which a periodic wave repeats, is its spatial period. It is the distance between two successive corresponding wave points of the same phase, such as two neighbouring crests, troughs, or zero crossings, and it is a feature of both travelling waves and standing waves, as well as other spatial wave patterns. The spatial frequency is defined as the wavelength's inverse. The Greek letter lambda, or "wavelength," is frequently used to represent it. Additionally, sinusoidal wave envelopes, modulated waves, and waves created by the interference of several sinusoids are also commonly referred to as having a wavelength.

You measure an angle of 21. 1 when the light passes through a grating with 610 lines per mm. what is the wavelength of the light?

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consider two charges. one has a charge of +15.mC and the other a charge of -3.00mC. if the electrostatic force between these cha

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Answer:164.3metre

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A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 6.0 m, y = 10.0 m, and has vel

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Explanation:

Given that,

Position of the particle at t = 0,

$y=(6i+10j)\ m$

Velocity of the particle at t = 0

$u=(1i+6j)\ m$

Acceleration of the particle,

$a=(5i+7j)\ m/s^2$

Solution,

(a) Let v is the velocity at t = 10 s. Using the equation of kinematics as :

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$v=(51i+76j)\ m/s$

(b) Let y' is the position at t = 1 s. Again using second equation of kinematics as :

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

$y'=(6i+10j)+(1i+6j)1+\dfrac{1}{2}\times (5i+7j)1^2$

$y'=\dfrac{19}{2}i+\dfrac{39}{2}j$

(c) Magnitude of y',

$|y'|=\sqrt{(\dfrac{19}{2})^2+(\dfrac{39}{2})^2}$

|y'| = 21.69 meters

Direction of the y',

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A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a

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Answer:

a) fr = 266.92 N, fy = 1300 N, b) μ = 0.36

Explanation:

a) This is a balancing act.

Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive

-w x - W x₂ + R y = 0 (1)

usemso trigonometry to find distances

cos 60.08 = x / 7.5

x = 7.5 cos 60.08

x = 3.74 m

fireman

cos 60.08 = x₂ / 4

x2 = 4 cos 60

x2 = 2 m

wall support

sin 60.08 = y / 15

y = 15 are 60.08

y = 13 m

we substitute in equation 1

R y = w x + W x2

R = (w x + W x2) / y

R = (500 3.74 +800 2) / 13

R = 266.92 N

now let's write the expressions for the translational equilibrium

X axis

R -fr = 0

R = fr

fr = 266.92 N

Y Axis

Fy - w-W = 0

fy = 500 + 800

fy = 1300 N

b) ask the friction coefficient

the firefighter's distance is

cos 60.08 = x₃ / 9.00

x₃ = 9 cos 60

x₃ = 5.28 m

from equation 1

R = (w x + W x₃) / y

R = 500 3.74 + 800 5.28) / 13

R = 468.769 N

we saw that

fr = R = 468.769

The expression for the friction force is

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N = Fy

N = 1300 N

μ N = fr

μ = fr / N

μ = 468,769 / 1300

μ = 0.36

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A dog running at 10 m/s is 30m behind a rabbit moving at 5 m/s. when will the dog catch up with the rabbit assuming both their v

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The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.

<h3>What time will the dog catch the rabbit?</h3>

The time that the dog will catch up with the rabbit is given as follows:

Let the distance covered by the rabbit be x.

Distance covered by dog = x + 30

Time taken = distance/speed

The time taken will be the same T

Time taken by dog, T = (x + 30)/10
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Equating both times.

(x + 30)/10 = x/5

x = 30 m

Solving for T in equation (ii);

T = 30/5 = 6 minutes

In conclusion, time is obtained as a ratio of distance and speed.

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