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Bezzdna [24]
3 years ago
12

A car slows from 22m/s to 3.0m/a at a constant rate of 2.1m/s. How many seconds are required before the car is traveling at 3.0m

/s?
Physics
1 answer:
lilavasa [31]3 years ago
8 0

Answer:

t = 9.05 s

Explanation:

Given,

The initial velocity of the car, u = 22 m/s

The final velocity of the car, v = 3.0 m/s

The rate of change of speed of the car is 2.1 m/s,

                                      ∴ the acceleration, a = -2.1 m/s²

The acceleration and velocity is given by the relation

                                           a = (v - u)/t     m/s²

Therefore,

                                           t = (v-u)/a  s

Substituting the values in the above equation

                                          t = (3.0 - 22)/(-2.1)

                                            = 9.05 s

Hence, the time required before the car is traveling at 3.0m/s is t = 9.05 s

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Object A with a mass of 500 kilograms hits stationary object B with a mass of 920 kilograms. If the collision is elastic, what h
Vika [28.1K]
In elastic collision, both the kinetic energy and momentum are conserved. Conservation means that both the kinetic energy and momentum will have the same values before and after elastic collision.

<span>As the object A has low mass than object B. Hence upon collision, object B moves forward, while object A will move backward. So option "C" is correct. </span>

5 0
4 years ago
g 4. A student with a mass of m rides a roller coaster with a loop with a radius of curvature of r. What is the minimum speed th
photoshop1234 [79]

Answer:

minimum speed v=\sqrt({Fr}/m)}

Explanation:

Recall the formula for centripetal force;

Centripetal force is the force that is required to keep an object moving in circular part

     F=mv^{2} /r

where;

F=centripetal force

m=mass of object

r=radius of curvature

v= minimum speed

To find minimum speed make v the subject of formula;

   v=\sqrt({Fr}/m)}

4 0
3 years ago
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How much force is needed to stop a 80-kg football player if he decelerates at 5 m/s2
kiruha [24]
According to Newton's 2nd law of motion, F =ma = 80x5 = 400N.
6 0
4 years ago
What is the concentration of H+ ions at a pH = 8?
skelet666 [1.2K]

Taking into account the definition of pH and pOH, the concentration of H⁺ ions at pH=8 is 1×10⁻⁸ M and the concentration of OH⁻ ions at pH=8 is 1×10⁻⁶ M.

<h3>Definition of pH</h3>

pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.

The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or H₃O⁺:

pH= - log [H⁺]= - log [H₃O⁺]

<h3>Definition of pOH</h3>

Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

<h3 /><h3>Relationship between pH and pOH</h3>

The following relationship can be established between pH and pOH:

pOH + pH= 14

<h3>Concentration of H⁺ ions</h3>

In this case, pH=8. Replacing in the definition of pH:

8= - log [H⁺]

Solving:

<u><em>[H⁺]= 10⁻⁸= 1×10⁻⁸ M</em></u>

Finally, the concentration of H⁺ ions at pH=8 is 1×10⁻⁸ M.

<h3>Concentration of OH⁻ ions</h3>

Being pH= 8, pOH is calculated as:

pOH + 8= 14

pOH= 14 - 8

pOH= 6

Replacing in the definition of pOH the concentration of OH⁻ ions is obtained:

- log [OH⁻]= 6

Solving

<u><em>[OH⁻]= 10⁻⁶= 1×10⁻⁶ M</em></u>

Finally, the concentration of OH⁻ ions at pH=8 is 1×10⁻⁶ M.

Learn more about pH and pOH:

brainly.com/question/16032912

brainly.com/question/13557815

#SPJ1

7 0
2 years ago
4. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa
shusha [124]

Answer:

a. 192 m/s

b. -17,760 kPa

Explanation:

First let's write the flow rate of the liquid, using the following equation:

Q = A*v

Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:

a.

A1*v1 = A2*v2

pi * 0.02^2 * 12 = pi * 0.005^2 * v2

v2 = 0.02^2 * 12 / 0.005^2

v2 = 192 m/s

b.

To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)

P1 + d1*v1^2/2 = P2 + d1*v2^2/2

Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)

600000 + 1000*12^2/2 = P2 + 1000*192^2/2

P2 = 600000 + 72000 - 1000*192^2/2

P2 = -17760000 N/m2 = -17,760 kPa

The speed in the smaller part of the pipe is too high, the negative pressure in the second part means that the inicial pressure is not enough to maintain this output speed.

4 0
3 years ago
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