All categories

4 years ago

When 108 g of water at a temperature of 22.5?

1 answer:

3 0

Î”T for the first sample is the total samples final temp, minus the first sample's initial temp (47.9-22.5), so 25.4oC. Calculating q for the first sample as 108g x 4.18 J/g C x 25.4oC = 11466.58 Joules Figuring that since the first sample gained heat, the second sample must have provided the heat, so doing the calculation for the second sample, I used q=mCÎ”T 11466.58 Joules = 65.1g x 4.18 J / g C x Î”T 11466.58/(65.1gx4.18)=Î”T Î”T=42.14oC So, since second sample lost heat, it's initial temperature was 90.04oC (47.9oC final temperature of mixture + 42.14oC Î”T of second sample).

You might be interested in

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

A spaceship approaches the earth with a speed 0.50c. A passenger in the spaceship measures his heartbeat as 9- beats per minute.

JulijaS [17]

Answer:

D) 61 beats per minute

Explanation:

5 0

3 years ago

A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 13° above the horizontal. (a) if

Blababa [14]

Answer: Therefore, x component: Tcos(24Â°) - f = 0 y component: N + Tsin(24Â°) - mg = 0 The two equations I get from this are: f = Tcos(24Â°) N = mg - Tsin(24Â°) In order for the crate to move, the friction force has to be greater than the normal force multiplied by the static coefficient, so... Tcos(24Â°) = 0.47 * (mg - Tsin(24Â°)) From all that I can get the equation I need for the tension, which, after some algebraic manipulation, yields: T = (mg * static coefficient) / (cos(24Â°) + sin(24Â°) * static coefficient) Then plugging in the values... T = 283.52. Reference https://www.physicsforums.com/threads/difficulty-with-force-problems-involving-friction.111768/

7 0

3 years ago

A solid 200-g block of lead and a solid 200-g block of copper are completely submerged in an aquarium filled with water. Each bl

finlep [7]

Answer:

B. The buoyant force on the copper block is greater than the buoyant force on the lead block.

Explanation:

Given;

mass of lead block, m₁ = 200 g = 0.2 kg

mass of copper block, m₂ = 200 g = 0.2 kg

density of water, ρ = 1 g/cm³

density of lead block, ρ₁ = 11.34 g/cm³

density of copper block, ρ₂ = 8.96 g/cm³

The buoyant force on each block is calculated as;

$F_B = mg(\frac{density \ of \ fluid}{density \ of \ object} )$

The buoyant force of lead block;

$F_{lead} = 0.2*9.8(\frac{1}{11.34} )\\\\F_{lead} = 0.173 \ N$

The buoyant force of copper block

$F_{copper} = 0.2*9.8(\frac{1}{8.96})\\\\F_{copper} = 0.219 \ N$

Therefore, the buoyant force on the copper block is greater than the buoyant force on the lead block

3 0

3 years ago

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

GuDViN [60]

Answer:

E/4

Explanation:

The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

Where;

E is the electric field

σ is the surface charge density

ε₀ is the electric constant.

Formula to calculate σ is;

σ = Q/A

Where;

Q is the total charge of the sheet

A is the sheet's area.

We are told the elastic sheet is a square with a side length as d, thus ;

A = d²

So;

σ = Q/d²

Putting Q/d² for σ in the electric field equation to obtain;

E = Q/(2ε₀d²)

Now, we can see that E is inversely proportional to the square of d i.e.

E ∝ 1/d²

The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.

From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;

E_new = E/4

3 0

3 years ago