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3 years ago

4. Which of the following is the term used to describe a body's resistance to a change in motion?

Physics

1 answer:

Alborosie3 years ago

5 0

Inertia ..................

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A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi

natali 33 [55]

Answer:

$v_1 = \ 5.196 \frac{m}{s}$

$v_2 = 3 \frac{m}{s}$

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

$\vec{p}_i = \vec{p}_f$

where the suffix i means initial, and the suffix f means final.

The initial momentum will be:

$\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}$

as the second puck is initially at rest:

$\vec{v}_{2_i} = 0$

Using the unit vector $\vec{i}$ pointing in the original line of motion:

$\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}$

$\vec{p}_i = 0.70 \ kg \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0$

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

So:

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f$

$\vec{p}_f = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

So, our velocity vectors will be:

$\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )$

$\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

We got

$\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}$

$4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \ v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ ) + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

So, we got the equations:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ v_1 \ cos(30 \°) + 0.7 \ kg \ v_2 \ cos(-60 \°)$

and

$0 = 0.7 \ kg \ v_1 \ sin(30 \°) + 0.7 \ kg \ v_2 \ sin(-60 \°)$ .

From the last one, we get:

$0 = 0.7 \ kg \ ( v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°) )$

$0 = v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°)$

$v_1 \ sin(30 \°) = - \ v_2 \ sin(-60 \°)$

$v_1 = \ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) }$

and, for the first one:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ ( v_1 \ cos(30 \°) + v_2 \ cos(60 \°) )$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = (\ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = v_2 (\ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + cos(60 \°)$

$6 \ \frac{m}{s} = v_2 * 2$

so:

$v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}$

and

$v_1 = \ 3 \frac{m}{s} \ \frac{sin(60 \°)}{ sin(30 \°) }$

$v_1 = \ 5.196 \frac{m}{s}$

3 0

3 years ago

An object is falling from a height of 7.5 meters. At what height will it’s velocity be 7 meters/second?

Nady [450]

One of the equations of gravity is this:
${v}^{2} = {u}^{2} + 2gh$
Where v = final velocity which is 7m/s
u = initial velocity which is 0 for objects falling from a height
g = acceleration due to gravity and it is approximately 10m/s^2. It's a constant so pretty much remember this number. It's positive since the work being done is caused by gravity (in other words, it's falling down). It can also be negative if the work being down is against gravity (in other words, it's going up)
h = height of object

Substitute for the values and you should have something like this
${7}^{2} = {0}^{2} + 2 \times 10 \times h$
$49 = 0 + 20h$
$h = \frac{49}{20}$
$h = 2.5m$

6 0

3 years ago

A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height

rewona [7]

Answer:

$W=17085KJ$

Explanation:

From the question we are told that:

Height $H=16m$

Radius $R=3$

Height of water $H_w=9m$

Gravity $g=9.8m/s$

Density of water $\rho=1000kg/m^3$

Generally the equation for Volume of water is mathematically given by

$dv=\pi*r^2dy$

$dv=\frac{\piR^2}{H^2}(H-y)^2dy$

Where

y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

$dw=(pdv)g (H-y)$

Substituting dv

$dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)$

$dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy$

Therefore

$W=\int dw$

$W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy$

$W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)$

$W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0$

$W=3420.84*0.25[2401-65536]$

$W=17084965.5J$

$W=17085KJ$

4 0

3 years ago

A cat has a mass of 4.6 kg. What is its weight on Earth's surface?

andreyandreev [35.5K]

<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ Earth's gravity is approximately 9.81

weight = mass x gravity

weight = 4.6 x 9.81

weight = 45.126

Answer is B. 45N

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

4 0

3 years ago

Read 2 more answers

Whats it called when a force can change the shape of an object?

alekssr [168]

Force applied causes deformation in the object. It changes the relative positions of constituent particles in the crystal lattice.
As soon as that happens, the interatomic or intermolecular forces come into play and they, tend to restore the solid back to it's original shape.

This restoring force per unit area is called Stress . When external forces are removed, the internal forces tend to restore the solid back.
This property is called Elasticity .

However, no material is perfectly elastic and what happens is that, the body is not able to restore itself completely.

3 0

3 years ago

Read 2 more answers