Exercise 3 In an equation y mx+c; y and x have dimensions of length and c is constant. What are the dimensions of m'
A. Mass
B. Length
C. Time
D. m is Dimensionless

Answer:

I. The change in temperature of the object.

II. The initial volume of the object.

Explanation:

Generally, the volume of a solid object tends to increase as its temperature increases and this phenomenon is known as thermal expansion.

Hence, the quantities which determine how large the change in volume of a solid object is includes;

I. The change in temperature of the object.

II. The initial volume of the object.

This ultimately implies that, when a solid object is heated, the atoms of the object vibrate rapidly about their fixed points and thus, causing an increase in the volume of the object.

In conclusion, this scientific phenomenon known as thermal expansion is valid and true for all the three (3) states of matter;

Answer:

-22.1

Explanation:

1 / 4

Step 1. List horizontal (xxx) and vertical (yyy) variables

xxx-direction yyy-direction

t=\text?t=?t, equals, start text, question mark, end text t=\text?t=?t, equals, start text, question mark, end text

a_x=0a  

x

​  

=0a, start subscript, x, end subscript, equals, 0 a_y=-9.8\,\dfrac{\text m}{\text s^2}a  

y

​  

=−9.8  

s  

2

m

​  

a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction

\Delta x=255\,\text mΔx=255mdelta, x, equals, 255, start text, m, end text \Delta y=\text ?Δy=?delta, y, equals, start text, question mark, end text

v_x=v_{0x}v  

x

​  

=v  

0x

​  

v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript v_y=?v  

y

​  

=?v, start subscript, y, end subscript, equals, question mark

v_{0x}=120\,\dfrac{\text m}{\text s}v  

0x

​  

=120  

s

m

​  

v, start subscript, 0, x, end subscript, equals, 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction v_{0y}=0v  

0y

​  

=0v, start subscript, 0, y, end subscript, equals, 0

Note that there is no horizontal acceleration, so v_x=v_{0x}v  

x

​  

=v  

0x

​  

v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript. The time is the same for the xxx and yyy directions.

Also, the package has no initial vertical velocity.

Our yyy-direction variable list has too many unknowns to solve for \Delta yΔydelta, y directly. Since both the yyy- and xxx-directions have the same time ttt and horizontal acceleration is zero, we can solve for ttt from the xxx-direction motion by using equation:

\Delta x=v_xtΔx=v  

x

​  

tdelta, x, equals, v, start subscript, x, end subscript, t

Once we know ttt, we can solve for \Delta yΔydelta, y using the kinematic equation that does not include the unknown variable v_yv  

y

​  

v, start subscript, y, end subscript:

\Delta y=v_{0y}t+\dfrac {1}{2}a_yt^2Δy=v  

0y

​  

t+  

2

1

​  

a  

y

​  

t  

2

delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared

Hint #22 / 4

Step 2. Find ttt from horizontal variables

\begin{aligned}\Delta x&=v_xt \\\\ t&=\dfrac{\Delta x}{v_{0x}} \\\\ t&=\dfrac{255\,\text m}{120\dfrac{\text m}{\text s}} \\\\ &=2.125\,\text s \end{aligned}  

Δx

t

t

​  

=v  

x

​  

t

=  

v  

0x

​  

Δx

​  

=  

120  

s

m

​  

255m

​  

=2.125s

​  

Hint #33 / 4

Step 3. Find \Delta yΔydelta, y using ttt

Using ttt to solve for \Delta yΔydelta, y gives:

\begin{aligned}\Delta y&=v_{0y}t+\dfrac{1}{2}a_yt^2 \\\\ &=\cancel{ (0 )t}+\dfrac{1}{2}\left (-9.8\dfrac{\text m}{\text s^2}\right )\left(2.125\,\text s\right)^2 \\\\ &=-22.1\,\text m \end{aligned}  

Δy

​  

=v  

0y

​  

t+  

2

1

​  

a  

y

​  

t  

2

=  

(0)t

​  

+  

2

1

​  

(−9.8  

s  

2

m

​  

)(2.125s)  

2

=−22.1m

​  

Hint #44 / 4

The correct answer is -22.1\,\text m−22.1mminus, 22, point, 1, start text, m, end text.

Answer:

C

Explanation:

Ball Y will travel at a speed greater than 5 m/s in the opposite direction of travel as before the collision.

The ball Y  will travel at a speed greater than 5 m/s in the same direction of travel as before the collision.

The given parameters;

Apply the principle of conservation of linear momentum. The total momentum of the balls before collision must equal total momentum after collision.

[tex]m_xv_x_1 + m_yv_y_1 = m_x v_x_2 + m_yv_y_2\\\\m_xv_x_2 - m_xv_x_1 = -(m_yv_y_2 - m_yv_y_1)\\\\\Delta P_x = - \Delta P_y[/tex]

The impulse experienced by ball Y is calculated as follows;

[tex]\Delta P_y = m_yv_y_2- m_yv_y_1 = Ft = 40 \times \frac{1}{6} = 6.67 \ kgm/s[/tex]

The final speed of ball Y is calculated as follows;

[tex]m_yv_y_2 - m_yv_y_1 = 6.67 \\\\m_yv_y_2 = 6.67 + m_yv_y_1\\\\0.5(v_y_2) = 6.67 + (0.5 \times 5)\\\\0.5(v_y_2) = 9.17\\\\v_y_2 = \frac{9.17}{0.5} \\\\v_y_2 = 18.34 \ m/s[/tex]

Thus, the ball Y  will travel at a speed greater than 5 m/s in the same direction of travel as before the collision.

Learn more here:https://brainly.com/question/22698801

The magnitude of the vertical component of the projectile's initial velocity is 200 m/s × sin 30°.

The diagrammatic representation of the velocity of the projectile can be seen in the attached image below.

From the diagram, let consider the ΔOAP where Vector OP makes an ∠θ = 30° to the horizontal x-axis.

where;

∴

Using trigonometric approach for ΔOAP;

[tex]\mathbf{sin\theta = \dfrac{AP}{OP}}[/tex]

[tex]\mathbf{AP =OP\times sin \theta}}[/tex]

AP = 200 × sin 30°

Learn more about projectile here:

https://brainly.com/question/20689870?referrer=searchResults

Answer: the total energy of the puck, ice surface, and surrounding air decreases to zero

Explanation:

A hockey puck slides across the ice and eventually comes to a stop because of friction between surface of the puck and ice surface.

Friction is a resistance to motion of the object. for example, when a body slides on horizontal surface in positive x direction, it has friction in negative x direction and that measure of friction is a frictional force. frictional force is directly proportional to the Normal(N). i.e. [tex]F_{fri}[/tex] ∝ N

[tex]F_{fri}[/tex] = μN where μ is called as coefficient of the friction. It is a dimensionless quantity.

When a body is kept on horizontal surface, its normal will be straight upward which is reaction of mg. i.e. N=mg.

Frictional force is equal to

[tex]F_{fri}[/tex] = μmg

When hockey puck slides across the ice, friction between surface of puck and surface of ice produces resistance to the motion of the puck, due to resistance puck slow down slowly and eventually come to a stop. Motion and frictional force are opposite to each other. we can calculated the exact value of frictional force when we know the coefficient of friction between puck and ice surface.

hence due to friction, puck come to a stop

To know more about frictional force, click :

https://brainly.com/question/30280752

#SPJ3

Answer:

Explanation:

The solution of the question cab e found in attachment below:

Answer:

maximum height = 0.1225 m

Explanation:

given data

H = 2m

t = 1 sec

solution

we consider here velocity of pot at lower side is u

and final velocity is v with acceleration a

time take is t/2

so

height h = u × [tex]\frac{t}{2}[/tex]  - 0.5 × g × [tex](\frac{t}{2})^2[/tex]     .................1

here

u = [tex]\frac{2}{t} \times ( h + \frac{gt^2}{8} )[/tex]

and

v = u +a t/2      .........................2

v = u + g t/2

v = [tex]\frac{2h}{t} + \frac{gt}{4} - \frac{gt}{2}[/tex]

v = [tex]\frac{2h}{t} - \frac{gt}{4}[/tex]

so that

maximum height is  = [tex]\frac{v^2}{2g}[/tex]

maximum height = [tex]\frac{(\frac{2h}{t} - \frac{gt}{4})^2}{2g}[/tex]

put here value of h and t

maximum height =  [tex]\frac{(\frac{2(2)}{1} - \frac{g(1)}{4})^2}{2g}[/tex]

maximum height = 0.1225 m

Answer:

Aluminum

Explanation:

promise

give  me brainlest plleaseee

Answer:aluminum

Explanation:

Answer:

51.04 joules

Explanation:

The movement by the gloves = 11cm

Force by the baseball = 464N

First of all we are going to convert cm to metres

11cm = 11/100 meters = 0.11m

The formula for workdone is given as:

W = f x d

W = workdone

F = force

D = distance

Workdone = 464 x 0.11

= 51.04 joules

The workdone by the ball is 51.04 joules

Thank you

Answer:

b) stopping the object

Explanation:

Friction always slows a moving object down. ... Friction can be a useful force because it prevents our shoes slipping on the pavement when we walk and stops car tyres skidding on the road. When you walk, friction is caused between the tread on shoes and the ground. This friction acts to grip the ground and prevent sliding

Answer:

(e) a and c above.

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

[tex] Momentum = Mass * Velocity [/tex]

The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.

Padded dashboards in cars are safer in an accident than non-padded ones because an occupant hitting the dashboard of an automobile car has an increased time of impact and a decreased impact force because the force or shock experienced is high and happens rapidly over a short period of time, thus, the occupant has less time and velocity while absorbing the momentum of the car in the course of the collision.

Answer:

The BTU, or British thermal unit, is actually a measure of heat.