(6) Module 4 Study Guide

The x vector component of a displacement vector has a magnitude of 86.2 m and points along the negative x axis. The y vector com

Flauer [41]

Answer:

(a) Magnitude of Vector = 207.73 m

(b) Direction = 65.48°

Explanation:

(a)

The formula to find out the magnitude of a resultant vector with the help of its x and y components is given as follows:

$Magnitude\ of\ Vector = \sqrt{d_{x}^{2} + d_{y}^{2}} = \sqrt{(86.2\ m)^{2} + (189\ m)^{2}}\\\\$

<u>Magnitude of Vector = 207.73 m</u>

(b)

For the direction of the vector we have the formula:

$Direction = tan^{-1}(\frac{y}{x})\\\\Direction = tan^{-1}(\frac{189\ m}{86.2\ m})\\\\$

<u>Direction = 65.48°</u>

4 0

3 years ago

10. A person with a mass of 15 kg is walking uphill at a velocity of 2 m/s. What is the walker's momentum?

marin [14]

Answer:

10. B. 30 kg.m/s

16. B. Elastic collision

20. Velocity

Explanation:

10. Momentum is simply the product of mass and velocity of an object. So it can be soved with the equation:

p = mv

Where:

p = momentum (kg.m/s)

m = mass (kg)

v = velocity (m/s)

Your problem gives you the following:

m = 15 kg

v = 2 m/s

p =?

So just fill in what you know:

p = mv

p = (15kg)(2m/s) = 30 kg.m/s

16. Elastic collision occurs when two objects collide and no loss in energy occurs after collision. Supposing that the balls are very bouncy, let's say they bumped into each other and they bounce-off of each other. If there is no loss in speed on both ends, that means that this is an elastic collision.

20. To differentiate vector and scalar quantity, just remember that vectors have direction. Both quanitites show magnitude, but vector quantities indicate the direction the magnitude is going. Among all the choices, it is only velocity that is considered as a VECTOR quantity.

3 0

3 years ago

oe finds that the temperature of a substance is 12 degrees Celsius. What does this tell Zoe about the substance? Its internal en

Roman55 [17]

Its internal energy is less than 12 degrees

5 0

3 years ago

Read 2 more answers

a ball of mass 15 kg moving at +20 m/s collides head-on with another ball of mass 10 kg moving at -15 Ms. if after collision the

Karo-lina-s [1.5K]

Given

m1 = 15kg

vi1 = +20 m/s

vf1 = +5 m/s

m2 = 10 kg

vi2= -15 m/s

vf2 = ?

Procedure

Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant.

$\begin{gathered} m_1v_{i1}+m_2v_{i2}=m_1v_{f2}+m_2v_{f2} \\ 15kg\cdot20m/s+10\operatorname{kg}\cdot-15m/s=15\operatorname{kg}\cdot5m/s+10\operatorname{kg}\cdot v_{f2} \\ 150-75=10v_{f2} \\ v_{f2}=75/10 \\ v_{f2}=7.5\text{ m/s} \end{gathered}$

The speed of the 10kg Ball is 7.5m/s

8 0

1 year ago

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.

(c) The frequency of the oscillation is given by :

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

Hence, this is the required solution.

7 0

4 years ago