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loris [4]
1 year ago
13

I need help with problem C, finding the area of the accel. v time graph

Physics
1 answer:
erastova [34]1 year ago
7 0

c) To determine the velocity of the motion at the given time, consider that the velocity is also the area under the line of the graph a vs t.

Between t=0s and t = 2 s, the area under the line is:

A1 = 2 X 2 = 4

Then, until t = 2s the velocity is 4 m/s

The area between t = 2s and t = 4 s is:

A2 = 2 x 0 = 0

The area between t = 4s and t = 5s, the area is:

A3 = 2 x 1 = 2

Now, consider that the velocity is the sum of the previous areas, but taking areas on the negative y-axis as negative ones. Then, you have:

A = A1 + A2 - A3 = 4 + 0 - 2 = 2

Hence, the velocity is 2 m/s

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Anne releases a stone from a height of 2 meters. She measures the kinetic energy of the stone at 9.8 joules at the exact point i
insens350 [35]
A. 0.5kg

To get this answer you need to follow the equation of KE=0.5*mv^2 
But we don't have the m part in the equation. So just plug in the numbers to see which works best, though I can tell you before we do that the answer would be a. 

As you may know, gravity, is a force of 9.8 m/s. And we want to get 9.8 Joules. So if we take a half a kg stone, release it at one meter, we get half of the normal gravity pull, 4.90 Joules. That means if we take half a kg stone and drop it at a doubled height, we get 9.8 Joules.

That is also to say that if we have a 1kg stone and drop it at one meter you will get the normal pull of gravity in Joules, 9.8J. 

Be careful though, this does not mean if you drop a 1kg stone and a .5 kg stone the 1kg will hit first. This simply means that the 1kg stone will have twice the Joules that the .5kg stone has.
7 0
3 years ago
Read 2 more answers
For the potential , determine the number of bound states?A)6 b)4 c) 3 d) 1
s2008m [1.1K]

Do you see that blank, open space after the word "potential ..." ?

There's supposed to be a number there that actually tells us the value of the potential.  Without that number ... and a lot more description of the whole scenario here ... there's no possible answer to the question.

7 0
3 years ago
A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which
julia-pushkina [17]

Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = 11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = 390\ W/m.^{\circ}C

Conduction of heat from the rod per second is given by:

q = \frac{KA\Delta T}{L}

where

\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C = temperature difference between the two ends of the rod.

Thus

q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s

Now,

To calculate the mass, M of the ice melted per sec:

M = \frac{q}{L_{w}}

where

L_{w} = Latent heat of fusion of water = 333 kJ/kg

M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g

5 0
3 years ago
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
3 years ago
A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant
vovangra [49]

Answer:

emf induced in the loop, at the instant when 9.0s have passed = 1.576 * 10 ⁻² V.

Direction is counter clockwise.

Explanation:

See attached pictures.

6 0
3 years ago
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