How many joules of heat would be required to heat 0.5 kg of aluminum by 2 kelvin

A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The

luda_lava [24]

Answer:

the rotational inertia of the cylinder = 4.85 kgm²

the mass moved 7.942 m/s

Explanation:

Formula for calculating Inertia can be expressed as:

$I =\frac{1}{2}mR^2$

For calculating the rotational inertia of the cylinder ; we have;

$I = \frac{1}{2}m_pR^2$

$I = \frac{1}{2}*10.53*(0.96)^2$

$I=5.265*(0.96)^2$

$I=4.852224$

I ≅ 4.85 kgm²

mg - T ma and RT = I ∝

T = $\frac{Ia}{R^2}$

$a = \frac{g}{1+\frac{I}{mR^2}}$

$a = \frac{9.8}{1+\frac{4.85}{3.9*(0.96)^2}}$

a = 4.1713 m/s²

Using the equation of motion

$v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s$

3 0

3 years ago

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A heat engine performs (245 + A) J of work in each cycle while also delivering (142 + B) J of heat to the cold reservoir. Find t

Ganezh [65]

Answer:

The value is $\eta = 54.4 \%$

Explanation:

From the question we are told that

The work input is $W = ( 245 + A ) \ J$

The heat delivered is $Q = (142 + B) \ J$

The value of A is A = 14

The value of B is B = 72

Generally the efficiency of the heat engine is mathematically represented as

$\eta = \frac{W}{Q_t}$

Here $Q_t$ is the total out energy produce by the heat engine and this is mathematically represented as

$Q_t= Q + W$

=> $Q_t= 245 + A + 142 + B$

=> $Q_t= 390 + A+B$

So

$\eta = \frac{245 + A }{390 + A+ B}$

=> $\eta = 0.544$

=> $\eta = 0.544 *100$

=> $\eta = 54.4 \%$

5 0

3 years ago

Match the type of boundary with it's characteristic

vodomira [7]

Answer:

Transform= not destroyed or created

Divergent= crust created

Convergent= crust destroyed

Explanation:

The plates move in the opposite or away from each other at a transforming plate boundary. The two platform borders are not produced or destroyed in this case. As both plates converge on each other and thus destroy the plates for converging plate boundaries. When the plate is divergent, both plates shift away from each other by opening up and solidification for a new crust.

3 0

3 years ago

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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

8 0

4 years ago

A reasonable estimate for the height of the walls in an ordinary american home is?

Olegator [25]

A reasonable estimate for the height of the walls in an ordinary American home is 2.5m.

<h3>What is wall estimation?</h3>

Building gauge is of two sorts one is known as Rough Cost Estimate. This gauge won't depict the specific expense of the structure, however it can provide you with the rough expense worth of the structure that will help adequately in overseeing cash for the structure. Harsh Cost Estimate is directed in various ways for various sort of structures. The rough complete wall length is found in running meters in this framework and this all out length accumulated by the rate per running meter of the wall gives an actually solid cost. Assessment hypothesis is a part of insights that arrangements with assessing the upsides of boundaries in light of estimated experimental information that has an irregular part. The boundaries depict a fundamental actual setting so that their worth influences the circulation of the deliberate information.

Learn more about wall estimation, visit

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8 0

1 year ago