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3 years ago

A boiler is being used to heat water. The graph shows the temperature of

Physics

1 answer:

Vinvika [58]3 years ago

4 0

Answer:

a. Liquid state

b. At point R. The physical state of water at its boiling point temperature of 100 degree Celsius will be both liquid state as well as gaseous state.

c. at 110 degrees it is in gaseous state

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What distance will a vehicle travel before coming to a complete stop from a speed of 70 mph, (a) When the vehicle is traveling o

OLga [1]

Answer:

(a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be $723.91>SSD>620.2$

(c), The SSD will be 910.5 ft.

Explanation:

Given that,

Speed = 70 mph

Suppose, a perception reaction time of 2.5 sec and the coefficient of friction is 0.35

We need to calculate the stopping sight distance

Using formula of SSD

$SSD=1.47\times v\times t+\dfrac{v^2}{30\times(f\pm g)}$

Where, v = speed of vehicle

t = perception reaction time

f = coefficient of friction

g = gradient of road

(a). If the gradient of road is zero.

Then, the stopping sight distance will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35)}$

$SSD=723.9\ ft$

(b-1). If the gradient of road is 0.1

Then, the stopping sight distance will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35+0.1)}$

$SSD=620.2\ ft$

(b-2). If the grade continuously decrease then the SSD will be increase.

But if the grade is increase then the SSD will be decrease and for flat grade the SSD will be more.

So, The SSD will be $723.91>SSD>620.2$

(c). When the vehicle is traveling downhill on a roadway of constant grade then the vehicle take will be more SSD

So, The SSD will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35-0.1)}$

$SSD=910.5\ ft$

Hence, (a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be $723.91>SSD>620.2$

(c), The SSD will be 910.5 ft.

7 0

3 years ago

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

3 years ago

The acceleration of an object as a function of time is given by a(t) = (1.00 m/s2)t2. If displacement of the object between time

jolli1 [7]

not enough information is given to determine the velocity of the object at time to=0.00s

3 0

3 years ago

While doing a lab, a student found the density of a piece of pure aluminum to be 2.85 g/cm3. The accepted value for the density

bezimeni [28]

The equation for percent error is

% Error = $100*|Experimental-Theoretical|/Theoretical$

Our experimental is 2.85g/cm^3 and the accepted is 2.7g/cm^3

Thus our % Error = 5.555%

3 0

3 years ago

Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v

wlad13 [49]

1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we call $v_f$ the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
$m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f$ (1)
where
$m_A=7 kg$ is the mass of ball A
$m_B=2 kg$ is the mass of ball B
$v_{Ai}=6 m/s$ is the initial velocity of ball A
$v_{Bi}=-12 m/s$ is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find $v_f$ , we find that the final velocity of the balls is
$v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s$
and the positive sign means the two balls are going to the right.

2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
$m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}$
$\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2$
where
$v_{fA}$ is the final velocity of ball A
$v_{fB}$ is the final velocity of ball B

If we solve simultaneously the two equations, we find:
$v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s$
$v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s$
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
$p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s$
and the total momentum after the collision is:
$p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s$

3 0

3 years ago