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3 years ago

When divided by △t, (vf - vi) is used to determine which characteristic? A.Speed B.Direction C. Displacement D.Acceleration

Physics

1 answer:

Keith_Richards [23]3 years ago

6 0

Answer:

D.Acceleration

Explanation:

The acceleration of an object is defined as:

$a=\frac{v_f -v_i}{\Delta t}$

where

vf is the final velocity of the object

vi is the initial velocity of the object

$\Delta t$ is the time taken for the change in velocity to occur

Therefore, from the formula, we see that if we divide (vf - vi) by △t, we get the acceleration.

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What is the internal energy of 2.00 mol of diatomic hydrogen gas (H2) at 35°C?

djyliett [7]

As you mentioned, we will use Equipartition Theorem.
H2 has 5 degrees of freedom; 3 translations and 2 rotation
Therefore:
Internal energy = (5/2) nRT
You just substitute in the equation with the values of R and T and calculate the internal energy as follows:
Internal energy = (5/2) x 2 x 8.314 x 308 = 32.0089 x 10^3 J

4 0

3 years ago

2. A person applies a force of 66 N to a fridge as they push it across the length of a standard tennis court. So far today, the

Lubov Fominskaja [6]

Answer:

$P=39.2205\, watt$

$E=374.948 \,cal$

Explanation:

Given that:

force applied, $F=66\,N$

displacement, $s=23.77\,m$ (length of a tennis court)

time taken for pushing, t = 40 s

Since, work is given by:

$W=F.s$

$W=66\times 23.77$

$W=1568.82\,J$

Now, power is given as:

$P=\frac{W}{t}$

$P=\frac{1568.82}{40}$

$P=39.2205 \,watt$

Calories consumed is:

$E= 1568.82\times 0.239$

$E=374.948\, cal$

3 0

4 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$