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2 years ago

A blue puck has a velocity of 3i –4j m/s. Its mass is 20 kg. What is its momentum?

Physics

2 answers:

damaskus [11]2 years ago

6 0

P = m * v
v = {3i - 4j} = square root (3^2 + 4^2) = 5
P = 20 * 5
P = 100 kg m/s

Tom [10]2 years ago

3 0

Magnitude:

√(3² + 4²) = √(9 + 16) = √25 = 5

Momentum = (mass) x (velocity) = (20kg x 5m/s) = 100 kg-m/s

Direction:

3i - 4j is in the direction tan⁻¹(-4/3) = 53.1 south of east.

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A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of

maks197457 [2]

Answer:

$5.6\times 10^{-8}\ Ohm.m$

Explanation:

Resistivity is given by $\rho=\frac {AR}{L}$ where A is cross-sectional area, R is resistance, L is the length and $\rho$ is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

$\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m$

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2 years ago

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to

Salsk061 [2.6K]

Explanation:

$1.2 \mathrm{N} ; 2 \mathrm{N}$

2. $200 \mathrm{N} ; 200 \mathrm{N}$

4. $2 \mathrm{N} ; 2 \mathrm{N} ; 4 \mathrm{N}$

$5.2 \mathrm{N} ; 2 \mathrm{N} ; 2 \mathrm{N}$

$6.2 \mathrm{N} ; 2 \mathrm{N} ; 3 \mathrm{N}$

$8.200 \mathrm{N} ; 200 \mathrm{N} ; 5 \mathrm{N}$

In only the above cases (i.e 1,2,4,5,6,8 ) the object possibly moves at a constant velocity of $256 \mathrm{m} / \mathrm{s}$

You should have noticed that the sets of forces applied to the object are the same asthe ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes nodistinction between the state of re st and the state of moving at a constant velocity(even a high velocity).

In both cases, the net force applied to the object must equal zero.

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2 years ago

For the first 10 seconds a squirrel runs 3 m/s to look for an acorn. The next 5 seconds he eats an acorn that he finds. Afterwar

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Distance covered by the squirrel to look for an acorn :

d = ( 3 m/s ) × 10 s = 30 m.

Time taken to eat an Acron is 5 seconds.

Time taken to cover distance of 30 m with 2 m/s speed is :

$T=\dfrac{30}{2}\ s= 15 \ s$

Therefore, total time take to get back to where he started is ( 10+5+15 ) = 30 s.

Hence, this is the required solution.

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2 years ago

Two objects (45.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h

Oksanka [162]

a) The acceleration of the objects is $3.56 m/s^2$

b) The tension in the string is 280.8 N

Explanation:

We start by writing the equations of motion for the two masses attached to the pulley.

For the heavier mass, we have:

$m_1 g - T = m_1 a$ (1)

where

$m_1 = 45.0 kg$ is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

T is the tension in the string

a is the acceleration of the system (here we assumed that the heavier mass accelerates downward)

For the lighter mass, we have

$T-m_2 g = m_2 a$ (2)

where

T is the tension in the string

$m_2 = 21.0 kg$ is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

a is the acceleration of the system (here we assumed that the lighter mass accelerates upward)

From (1) we get

$T=m_1g - m_1 a$

And substituting into (2),

$(m_1 g - m_1 a)-m_2 g = m_2 a\\(m_1 -m_2)g = (m_1+m_2)a\\a=\frac{m_1 - m_2}{m_1+m_2}g=\frac{45-21}{45+21}(9.8)=3.56 m/s^2$

From the previous part of the problem we got an expression for the tension in the string:

$T=m_1g - m_1 a$

Where we have

$m_1 = 45.0 kg$

$g=9.8 m/s^2$

$a=3.56 m/s^2$ is the acceleration, found in part a)

Susbtituting, we find

$T=(45.0)(9.8)-(45.0)(3.56)=280.8 N$

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2 years ago

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"If one increases the force on an object, its acceleration increases too because the push it feels is greater"

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2 years ago

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