Describe what’s going on in position-time graphs 1 , 3, &5

a pool ball leaves a 0.60 meter high table with an initial high table with an initial horizontal velocity of 2.4m/s. predict the

timurjin [86]

Ball leaves the table of height 0.60 m

Initial speed of the ball is in horizontal direction which is 2.4 m/s

Now here we can say that time taken by the ball to hit the ground will be given by kinematics equation in vertical direction

$\Delta y = v_y*t + \frac{1}{2}at^2$

here in y direction we can say

$\Delta y = 0.60 m$

$v_y = 0$

$a = 9.8 m/s^2$ due to gravity

now from above equation we have

$0.60 = 0 + \frac{1}{2}*9.8*t^2$

$t = 0.35 s$

Now in the same time the distance that ball move in horizontal direction is given as

$d = v_x* t$

given that

$v_x = 2.4 m/s$

$d = 2.4 * 0.35$

$d = 0.84 m$

So it will move a total distance of 0.84 m where it will land

5 0

3 years ago

A vector has what and direction

galben [10]

A vector Quantity has MAGNITUDE and direction, so the ‘what’ is Magnitude.

8 0

4 years ago

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In lecture, Dr. Selby used atmospheric air pressure to crush a can, by sucking the air out of it. (a) Calculate how many Newtons

Ahat [919]

Answer:

a) F1 = 1999.8 N , F2 = 4545 N , F3 = 2778 N , c) the cans do not collapse because the pressure is applied on both sides

Explanation:

Let's use the pressure equation

P = F / A

Suppose we have atmospheric pressure 1.01 10⁵ Pa

Let's calculate the area of the can that is a parallelepiped

Length L = 25 cm

width a = 18 cm

high h = 11 cm

Side area A = h a

A = 11 18

A1 = 198 10⁻⁴ m²

Lid area

A2 = L a

A2 = 25 18

A2 = 450 10⁻⁴ m²

Other side area

A3 = L h

A3 = 25 11

A3 = 275 10⁻⁴ m²

Now let's calculate the force on these sides

Side 1

F1 = P * A1

F1 = 1.01 10⁵ 198 10⁻⁴

F1 = 1999.8 N

Side 2

F2 = P A2

F2 = 1.01 10⁵ 450 10⁻⁴

F2 = 4545 N

Side 3

F3 P A3

F3 = 1.01 10⁵ 275 10⁻⁴

F3 = 2778 N

We see that the force is greater on side 2 which is where the can should collapse

b) To compare the previous forces we must use the concept of density, in general the cans are made of aluminum that has a density of 2700 kg / m3

d = m / V

m = d * V

V = L a h

V = 0.25 0.18 0.11

V = 0.00495 m3

m = 2700 0.00495

m = 13.4 kg

This is the maximum weight, because much of the volume we calculate is air that has a much lower density

W = 13.4 * 9.8

W = 131.3 N

Let's make the comparison by saying the two magnitudes

Side 1

F1 / W = 1999.8 / 131.3

F1 / W = 15.2

Side 2

F2 / W = 4545 / 131.3

F2 / W = 34.6

Side 3

F3 / W = 2778 / 131.3

F3 / W = 21.2

c) the cans do not collapse because the pressure is applied on both sides: outside and inside, so the net force is zero on each side.

7 0

4 years ago

A physical science test book has a mass of 2.2 kg what is the weight on the earth

timofeeve [1]

Answer:

21.582 Newtons

Explanation:

The weight of an object on Earth is equal to W=9.81M where M is the mass of the object and the acceleration due to gravity is 9.81 m/s^2. Therefore, the weight of the book on the earth is W=9.81(2.2)=21.582N, or 21.582 Newtons

3 0

3 years ago

A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the

MissTica

Answer : The value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

$q=n\times c_v\times \Delta T$

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

$\Delta T$ = Change in temperature = 15.7 K

$c_v$ = heat capacity at constant volume of $N_2$ (diatomic molecule) = $\frac{5}{2}R$

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

$q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K$

$q=286.2J$

Now we have to calculate the change in internal energy of the system.

$\Delta U=q+w$

As we know that, work done is zero at constant volume. So,

$\Delta U=q=286.2J$

Therefore, the value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

8 0

4 years ago

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