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kvv77 [185]
3 years ago
13

Moving from 0m/s to 25m/s in 8.0s equals an average acceleration of...

Physics
2 answers:
Verdich [7]3 years ago
5 0

Moving from 0m/s to 25m/s in 8.0s equals an average acceleration of 3.125 m/s^2. You get this through the following steps:

25-0/8-0 = 3.125 m/s^2

since the average acceleration (a) = change in vvelocity over change in time


Answer: 3.125 m/s^2


Shoutout to: @Mr. Toto




sad!; live


arlik [135]3 years ago
4 0
<h2>Answer:</h2>

Average acceleration = 3.124 m/s²

<h3>Explanation:</h3>

Given data:

Initial velocity = 0 m/s

Final velocity = 25 m/s

Time = 8.00 s

Average acceleration = ?

Solution:

Average acceleration = (Final velocity - Initial velocity)/time

By putting values in above formula:

Average acceleration = (25 - 0)/8

                                  = 25/8

                                  = 3.125 m/s²

Hence an object moving from 0m/s to 25m/s of velocity in 8.0s has an average acceleration of 3.125 meter per second square. Acceleration is the rate of change of velocity. And average acceleration is equal to change in final velocity and initial velocity divided by time.


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The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?
Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours.  So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes.  So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = <em>500 g</em> .

============================================

<u>Another way</u>:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = <em>500 g </em>.

5 0
3 years ago
A piece of warm concrete is placed in a cold-water tank, and energy flows between the concrete and the water. Which way does the
sammy [17]

Answer:

Heat flows from hot to cold objects. When a hot and a cold body are in thermal contact, they exchange heat energy until they reach thermal equilibrium, with the hot body cooling down and the cold body warming up. This is a natural phenomenon we experience all the time.

Explanation:

3 0
2 years ago
(10p+15,15-10q)=(25,5)​
Harrizon [31]
63783626736377474737377447
8 0
2 years ago
One particular descent goes from 2100m to 1600m. Assuming work done against friction is 90% of the potential energy change of th
alisha [4.7K]

Answer:

1/2 M V^2 = .1 M g H       where 10% of PE goes into KE

V^2 = .2 g H = .2 * 9.8 * (2100 - 1600) = 980 m^2 / s^2

V = 31.1 m/s       increase in speed during descent

1 km / hr = 1000 m / 3600 sec = .278 m/s

V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr

7 0
2 years ago
A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 94.8 g of water at 60.4°C with 94.8
madreJ [45]

Answer:

396.65 JC⁻¹

Explanation:

m_{a} = mass of water added to calorimeter = 94.8 g

T_{ai} = initial temperature of the water added = 60.4 C

c_{w} = specific heat of water = 4.184 Jg⁻¹C⁻¹

m_{c} = mass of water available to calorimeter = 94.8 g

T_{ci} = initial temperature of the water in calorimeter = 22.3 C

T_{f} = final equilibrium temperature = 35 C

Q_{C} = Heat gained by calorimeter

Using conservation of heat

Heat gained by calorimeter = Heat lost by water added - heat gained by water in calorimeter

Q_{C} = m_{a} c_{w} (T_{ai} - T_{f}) -  m_{c} c_{w} (T_{f} - T_{ci})

Q_{C} = (94.8) (4.184) (60.4 - 35) -  (94.8) (4.184) (35 - 22.3)

Q_{C} = 5037.4 J

T = Change in temperature of calorimeter

Change in temperature of calorimeter is given as

T = T_{f} - T_{ci}

T = 35 - 22.3

T = 12.7 C

Heat capacity of calorimeter is given as

c_{cm} = \frac{Q_{C}}{T}

c_{cm} = \frac{5037.4}{12.7}

c_{cm} = 396.65 JC⁻¹

7 0
3 years ago
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