All categories

2 years ago

(Repost) I need help with this physics question. Thanks in advance! Answer ASAP.

Physics

1 answer:

Kitty [74]2 years ago

3 0

The time taken for the tiny saliva to travel is 0.55 second.

The horizontal distance traveled at speed of 4 m/s is 2.2 m.

The horizontal distance traveled at speed of 20 m/s is 11 m.

<h3>Time of motion of the tiny saliva</h3>

The time taken for the tiny saliva to travel is calculated as follows;

h = vt + ¹/₂gt²

where;

v is initial vertical velocity = 0
g is the acceleration due to gravity

h = 0 + ¹/₂gt²

h = ¹/₂gt²

2h = gt²

t² = 2h/g

t = √(2h/g)

Substitute the given parameters and solve for time of motion;

t = √(2 x 1.5 / 10)

t = 0.55 second

<h3>Horizontal distance traveled at speed of 4 m/s</h3>

X = Vx(t)

X = (4 m/s)(0.55)

X = 2.2 m

<h3>Horizontal distance traveled at speed of 20 m/s</h3>

X = (20)(0.55)

X = 11 m

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

You might be interested in

Which statement can be supported by using a position-time grap

Phoenix [80]

Answer:

A negative slope results when an individual is moving away

Explanation:

6 0

3 years ago

Air in the amount of 2 lbm is contained in a well-insulated, rigid vessel equipped with a stirring paddle wheel. The initial sta

vladimir1956 [14]

Explanation:

It is known that energy balance relation is as follows.

$\Delta E_{system} = E_{in} - E_{out}$

Also, $W_{in} = \Delta U$

so, $W_{in} = mC_{v}(T_{2} - T_{1})$

According to the ideal gas equation,

$T_{2} = T_{1} \frac{P_{2}}{P_{1}}$

Putting the values into the above equation as follows.

$T_{2} = T_{1} \frac{P_{2}}{P_{1}}$

= $(520R) \frac{40psia}{30psia}$

= 693.3 R

Now, we will convert the temperature into degree Fahrenheit as follows.

693.3 - 458.67

= $234.63^{o}F$

From table A- $2E_{a}$

$C_{p}$ = 0.240 Btu/lbm R and $C_{v}$ = 0.171 Btu/lbm

Now, we will substitute the energy balance as follows.

$W_{in} = mC_{v}(T_{2} - T_{1})$

= $2 lbm \times 0.171 Btu/lbm R (693.3 - 520)$

= 59.3 Btu

Thus, we can conclude that final temperature of air is 59.3 Btu.

6 0

4 years ago

The take-up reel of a cassette tape has an average radius of 1.5 cm. Find the length of tape (in meters) that passes around the

Rudik [331]

Answer:

1.28 m

Explanation:

Given;

Radius, r = 1.5 cm = 0.015 m

Time, t = 19 s

Average angular speed = 4.5 rad/s

Consider a point when the tape is moving at a constant velocity along the circumference of the circular reel of radius r. The linear velocity v at this point is given by;

v = rω ----(1)

Where

v is the linear velocity of the circular motion

r is the radius of the reel

ω is the the angular velocity.

At a point the tap undergoes a linear motion before passing round the reel of the cassette. The linear velocity v at this point is given by;

v = L/t ----(2)

where;

v is the velocity of the linear motion

L is the length of the tape (distance covered by the tape)

t is the time taken

Equating equation(1) and equation (2)

L/t = rω

L = rωt

Substituting the given values,

L = 0.015 × 4.5 × 19

L = 1.2825 m

L = 1.28 m

4 0

4 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

The volume of the nucleus as compared to the volume of the atom is

velikii [3]

The nucleus of an atom is about 10-15 m in size; this means it is about 10-5 (or 1/100,000) of the size of the whole atom. A good comparison of the nucleus to the atom is like a pea in the middle of a racetrack. (10-15 m is typical for the smaller nuclei; larger ones go up to about 10 times that.)

4 0

3 years ago