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Alex17521 [72]
3 years ago
12

you apply the same amount of heat to five grams of water and five grams of aluminum the temperature of the aluminum increases mo

re than the temp of the water what can you conculde
Physics
2 answers:
Triss [41]3 years ago
6 0
I can conclude that the aluminum heats up faster then the water, since how they are made up is more different, each has a different compound to it, plus water is more denser<span />
blondinia [14]3 years ago
6 0

Answer:

As we know that heat given to a substance is given by the formula as

Q = m C\Delta T

now we know that

\Delta T = \frac{Q}{m C}

now we know that aluminium and water both are of same mass and both are given same amount of heat so the value of Q and m will be same.

So here we can say that change in temperature depends inversely on the specific heat capacity.

So here if more specific heat capacity then there will be less change in temperature.

so we can conclude that since temperature of aluminium increases more so the specific heat capacity of aluminium will be less than that of water.

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At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/
Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

4 0
3 years ago
What is the distance a truck can travel if it moves 7 m/s for 20 seconds?
artcher [175]

if it moves 7m/s, that means every second it goes 7m.  Now we just multiply by the time (20 seconds) and end up with 140m.

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3 years ago
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Air bags can be dangerous or fatal under certain circumstances. you should not sit closer than ______ inches from the steering w
pishuonlain [190]
<span>10 inches
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3 years ago
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One end of a spring with a force constant of k = 10.0 N/m is attached to the end of a long horizontal frictionless track and the
koban [17]

Answer:

-2.478

0.379

11.14

24.78

Explanation:

Angular frequency of spring in harmonic motion is given by?

ω = √(k/m)

ω = √(10/2.2)

ω = √4.54

ω = 2.13 s^-1

If at t=0 the mass is in negative amplitude (x = -A = -2.48 m) then we describe the position with negative cosine

x(t) = -A * cos(ωt)

x(t) = -2.48 * cos(2.13 * 1)

x(t) = -2.48 * 0.9993

x(t) = -2.478

Velocity and acceleration are 1st and 2nd derivative of position

b)

v(t) = Aω * sin(ωt)

v(t) = 2.48 * 2.13 * sin(2.13 * 1)

v(t) = 5.282 * sin2.13

v(t) = 5.282 * 0.03717

v(t) = 0.379 m/s

c)

a(t) = Aω^2 * cos(ωt)

a(t) = 2.48 * 2.12² * cos(2.13 * 1)

a(t) = 2.48 * 4.494 * cos2.13

a(t) = 11.15 * 0.9993

a(t) = 11.14 m/s²

d)

F = -k * x(t)

F = -10 * -2.478

F = 24.78 N

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Serga [27]
Due to the law of conservation of momentum, the force exerted on the mallet is equal and opposite to the force exerted on the ball, so the answer is C.
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