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4 years ago

Challenge Question (This is supposed to be hard, think critically and take it one step at a time):

Physics

1 answer:

aivan3 [116]4 years ago

8 0

Answer:

You must travel at an average speed of 67.06 m/s to be on time to the lecture.

Explanation:

From the question, the total distance from your house to the science lecture is 75 miles.

Also, you get halfway there before you stop for a gas, that is, you have covered half of 75 miles, which 37.5 miles and you also have to cover 37.5 miles to get to the science lecture.

After filling up, you only have 15 minutes before the lecture starts,

To determine how fast you must drive to be on time to the lecture,

we will determine the average speed you need to travel.

From

Average speed = Distance / Time

Distance = 37.5 miles (Convert to meters)

(NOTE: 1 mile = 1609.344 meters)

Hence, 37.5 miles = 37.5 × 1609.344 miles = 60350.4 meters

∴ Distance = 60350.4 meters

Time = 15 minutes (Convert to seconds)

(NOTE: 1 minute = 60 seconds)

Hence, 15 minutes = 15 × 60 seconds = 900 seconds

Now, from

Average speed = Distance / Time

Average speed = 60350.4 m / 900 s

Average speed = 67.06 m/s

Hence, you must travel at an average speed of 67.06 m/s to be on time to the lecture.

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3 years ago

Read 2 more answers

A 50-cm-long spring is suspended from the ceiling. A 330 g mass is connected to the end and held at rest with the spring unstret

lukranit [14]

Explanation:

Given that,

Length of the spring, l = 50 cm = 0.5 m

Mass connected to the end, m = 330 g = 0.33 kg

The mass is released and falls, stretching the spring by 30 cm before coming to rest at its lowest point. On applying Newton's second law, 10 cm below the release point, x = 15 cm

(a) When the mass is connected, the force of gravity is balanced by the force in spring.

$kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.15}\\\\k=21.56\ N/m$

(b) The amplitude of the oscillation will be 15 cm as it is half of the total distance travelled.

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{21.56}{0.33}} \\\\f=1.28\ Hz$

Hence, this is the required solution.

7 0

4 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago