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vivado [14]
3 years ago
10

A car traveling at a speed of 13 meters per second accelerates uniformly to a speed of 25 meters per second in 5.0 seconds. 11-

Calculate the magnitude of the acceleration of the car during this 5.0-second time interval.
Physics
1 answer:
Aleks04 [339]3 years ago
7 0

Answer:

a=2.4\ m/s^2

Explanation:

Given that,

Initial speed of a car, u = 13 m/s

Final speed of a car, v = 25 m/s

Time, t = 5 s

We need to find the acceleration of the car during this 5.0 second time interval. Let a is the acceleration. It can be calculated as :

a=\dfrac{v-u}{t}\\\\a=\dfrac{25-13}{5}\\\\=2.4\ m/s^2

So, the acceleration of the car is 2.4\ m/s^2.

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Pooping in my room and my room is upstairs and upstairs bathroom upstairs
diamong [38]

Answer:

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Explanation:

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7 0
2 years ago
A bag of ice cubes absorbs 149,000 J of heat, which causes its temperature to increase by 5.23 degrees celsius. what is the mass
Dima020 [189]

Answer:

14.3kg

Explanation:

Given parameters:

Quantity of heat = 149000J

Change in temperature = 5.23°C

specific heat of the ice = 2000J/kg°C

Unknown:

Mass of the ice in the bag = ?

Solution:

The heat capacity of a substance is given as:

            H = m c Ф

H is the heat capacity

m is the mass

c is the specific heat

Ф is the temperature change;

  since m is the unknown, we make it the subject of the expression;

                    m = H/ mФ

                  m = \frac{149000}{2000 x 5.23}  = 14.3kg

6 0
3 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
In the diagram, q1, q2, and q3 are in a straight line.
schepotkina [342]

Answer:

did anyone figure it out

Explanation:

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3 0
2 years ago
Would it be easier to pull the tablecloth out from under a glass of water or an
Aleksandr-060686 [28]
Glass as it is heavier and by newtons first law and inertia the greater the mass = more inertia(resistance to chabge in motion) thus the glass has a greater mass than an empty paper cup and thus has greater inertia. So it would be easier to leave it in place
3 0
2 years ago
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