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3 years ago

6

The graphs display velocity data Velocity is on the y-axis (m/s), while time is on the x-axis (s). Based on the graphs, which da

ta set represents constant acceleration?

2 answers:

N76 [4]3 years ago

6 0

Answer:

A

Explanation:

on edge

RUDIKE [14]3 years ago

5 0

Answer:

The first graph is showing the constant acceleration (1 m/s)

Explanation:

The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4

The last graph is showing constant velocity therefore there is no acceleration (a = 0)

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An arrow of mass 0.5kg so it has 25J of kinetic energy; find the speed of the arrow v = m/s

miss Akunina [59]

Answer:

10 m/s

Explanation:

Use the kinetic energy formula:

KE=(1/2)mv^2

I always remember it as Kevin is half-mad, and very square.

25J = (1/2)*0.5kg*(v^2)

50J = 0.5kg*(v^2)

100J = v^2

v = 10 m/s

Check it:

KE = (1/2)*0.5*(10^2)

KE = 25J

yep, it's right!

4 0

3 years ago

Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.)

Taya2010 [7]

Answer:

(a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

$P_{1}=\rho g h_{1}$

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

$P_{1}=\rho gh_{1}$ .....(I)

Pressure for second pipe,

$P_{2}=\rho gh_{2}$ .....(II)

From equation (I) and (II)

$P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

Put the value of P₁ and P₂

$\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)$

$2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2$ ....(III)

We know that,

The continuity equation

$v_{1}A_{1}=v_{2}A_{2}$

$v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})$

Put the value of v₂ in equation (III)

$2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2$

$2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2$

Here, $\dfrac{A_{1}}{A_{2}}=\gamma$

So, $2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)$

$v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Hence, (a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

8 0

3 years ago

A coffee-cup calorimeter is used to determine the heat of reaction for the reaction of compound A with compound B. A(aq) + B(aq)

Sergeeva-Olga [200]

Answer:

$\rm 1.64\times 10^{3}\; J$ .

Explanation:

Assume that his calorimeter is sufficiently effective, such that no heat had escaped to the surroundings. Heat from this solution would be absorbed by either

the solution, or
the coffee cup.

Temperature change: $33.637 - 23.722 = \rm 9.915\; ^\circ C$ .

<h3>Heat absorbed by the solution:</h3>

Only the specific heat capacity (per unit mass) of the solution is given. Both the mass of the solution and the temperature change will be required for determining the energy change. Start by finding the mass of the solution.

$m = \rho \cdot V = 2\times 16.10 \times 1.00 = \rm 32.10\; g$ .

Calculate the amount of heat absorbed from the specific heat:

$Q = c\cdot m \cdot \Delta T = 4.814\times 32.10 \times 9.915 = \rm 1335.80\; J$ .

<h3>Heat absorbed by the coffee cup:</h3>

The heat capacity of the coffee cup is given. Only the temperature change will be required for finding the amount of heat absorbed.

$Q = C\cdot \Delta T = \rm 30.7\times 9.915 = 304.391\; J$ .

<h3>Heat that this reaction produces</h3>

Find the sum of the two parts of heat. Round to three significant figures as in the heat capacity of the coffee cup and the density of the solution.

$\rm 1335.80+ 304.391 = 1.64\times 10^{3}\; J$ .

8 0

4 years ago

If the coolant in an air conditioner has a lower temperature than the air in a building, then ____.

solmaris [256]

Heat always flows from higher temperature to lower temperature. So option A. heat will flow from the air into the coolant is the correct answer.

8 0

4 years ago

Read 2 more answers

Pronouns in the nominative case may function as __________.

tresset_1 [31]

Pronouns in the nominative case may function as Subjects.

Like

We went to the game.
You played the game.
He/she/they watched the game.

8 0

3 years ago