1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic

How do I do these? My teacher didn’t show us how.

melisa1 [442]

Explanation:

Displacement is simply the change in position. So in the first part of problem 1, looking at the graph between 0 s and 2 s, the position changes from 0 m to -4 m. So the displacement is:

Δx = -4 m − 0 m

Δx = -4 m

Between 2 s and 4 s, the position stays at -4 m. The displacement is:

Δx = -4 m − (-4 m)

Δx = 0 m

Finally, between 4 s and 6 s, the position goes from -4 m to 6 m. The displacement is:

Δx = 6 m − (-4 m)

Δx = 10 m

The net displacement is the change in position from 0 s to 6 s:

Δx = 6 m − 0 m

Δx = 6 m

In the second part of problem 1, we have a velocity vs time graph.

Car 1 starts with 0 velocity and ends with a velocity of 6 m/s, so it is accelerating and constantly moving to the right.

Car 2 starts with a velocity of -6 m/s and ends with a velocity of 6 m/s. It is also accelerating, but first it is moving to the left, comes to a stop at t = 3 s, then moves to the right.

Car 3 starts with a velocity of 2 m/s and ends with a velocity of 2 m/s. So it is moving constantly to the right, but never speeds up or slows down.

We want to know when two of the cars meet. Unfortunately, this isn't as easy as looking for where the lines cross on the graph. We need to calculate their displacements. We can do this by finding the area under the graph (assuming all the cars start from the same point).

Let's start with Car 2. Half of the area is below the x-axis, and half is above. Without doing calculations, we can say the total displacement for this car is 0. This means it ends back up where it started, and that it never meets either of the other cars, both of which have positive displacements.

So we know Car 1 and Car 3 meet, we just have to find where and when. For Car 1, the area under the curve is a triangle. So its displacement is:

Δx = ½ t v(t)

where t is the time and v(t) is the velocity of Car 1 at that time. Since the line has a slope of 1 and y intercept of 0, we know v(t) = t. So:

Δx = ½ t²

Now look at Car 3. The area under the curve is a rectangle. So its displacement is:

Δx = 2t

When the two cars have the same displacement:

½ t² = 2t

t² = 4t

t² − 4t = 0

t (t − 4) = 0

t = 0, 4

t = 0 refers to the time when both cars are at the starting point, so t = 4 is the answer we're looking for. Where are the cars at this time? Simply plug in t = 4 into either of the equations we found:

Δx = 8

So Cars 1 and 3 meet at 4 s and 8 m.

7 0

3 years ago

How could you increase the number of snowmen destroyed by the sled sliding down the hill? SledAQ1 A. Increase the starting heigh

Dima020 [189]

To increase the number of snowmen destroyed, you would want more energy and all of the above would increase the energy of the system so D.

8 0

3 years ago

Read 2 more answers

Pls help me… I missed the whole lesson and I have no clue what to do-

galben [10]

Answer:

1) 50 facing towards the right

2) 150 facing right

3) 200 facing right

4) 0- no direction

5) 50- facing left

6) 50 facing right

Explanation:

forces in opposite directions and equal magnitudes counteract each other. in number 2 they face the same direction so they would just be added. in number 4 they oppose each other so would be subtracted

8 0

3 years ago

The sales price of a product is $2 per unit; the variable cost is $1 per unit; and fixed costs total $1,000. how many units must

irakobra [83]

Answer:

1000 units for breakeven

Explanation:

Let x be the number of units sold at breakeven.

The total sales at the point would be $2x.

Variable costs would be $1x and fixed costs are $1000.

Total costs are = $1x + $1000

At breakeven: Sales = Costs

Sales =m Costs

$2x = $1x + $1000

$1x = $1000

x = 1000 units.

At 1000 units the sales are equal to the costs ("breakeven").

4 0

2 years ago

Solve this physics for me please with steps

Mars2501 [29]

Answer:

The answers are located in each of the explanations showed below

Explanation:

a)

(i) Surface Tension: The tensile force that causes this tension acts parallel to the surface and is due to the forces of attraction between the molecules of the liquid. The magnitude of this force per unit of length is called surface tension.

σ = F/l [N/m]

where:

F = force [N]

l = length [m]

σ = Surface Tension [N/m]

(ii) Frequency is the number of repetitions per unit of time of any periodic event.

f = 1/T [1/s] or [s^-1] or [Hz]

where:

T = period [s] or [seconds]

f = frecuency [Hz] or [hertz]

(iii) Each of the units will be shown for each variable

v = velocity [m/s]

a = accelertion [m/s^2]

s = displacement [m]

$[\frac{m}{s} ]^{2} =[\frac{m}{s} ]^{2} + 2* [\frac{m}{s^{2} } ]*[m]\\$

$[\frac{m^2}{s^2} ] =[\frac{m^2}{s^2} ] + [\frac{m^{2} }{s^{2} } ]$

$[\frac{m^2}{s^2} ]$

b) To find the velocity we must derivate the function X with respect to t because this derivate will give us the equation for the velocity, it means:

$v=\frac{dx}{dt} \\v = 0.75*2*t+5*t$

$(i) X = 0.75*t^{2} +5*t+1\\X = 0.75*(4)^{2} +5*(4)+1\\X = 33 [m]$

ii) replacing in the derivated equation.

$v=1.5*(4)+5\\v=11[m/s]$

iii) the average velocity is defined by the expresion v = x/t

$v = \frac{x-x_{0} }{t-t_{0} } \\$

$x_{0}=0.75(2)^{2}+5(2)+1 \\ x_{0}=14[m]\\x=0.75(7)^{2}+5(7)+1\\x=72.75[m]\\t = 7 [s]t0= 2[s]Now replacing:[tex]v_{prom} = \frac{72.75-14}{7-2} \\v_{prom} = 11.75 [m/s]$

2

a) Pascal's principle or Pascal's law, where the pressure exerted on an incompressible fluid and in balance within a container of indeformable walls is transmitted with equal intensity in all directions and at all points of the fluid.

Therefore:

P1 = pressure at point 1.

P2 = pressure at point 2.

P1 = F1/A1

P2= F2/A2

$\frac{F_{1} }{A_{1} }=\frac{F_{2}}{A_{2} } \\F_{1}=A_{1}*(\frac{F_{2}}{A_{2} })$

b) One of the applications of the surface tension is the capillarity this is a property of liquids that depends on their surface tension (which, in turn, depends on the cohesion or intermolecular force of the liquid), which gives them the ability to climb or descend through a capillary tube.

Other examples of surface tension:

The mosquitoes that can sit on the water.

A clip on the water.

Some leaves that remain floating on the surface.

Some soaps and detergents on the water.

5 0

3 years ago

1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic​

1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic