What kind of thermal energy that flows between objects due to a difference in temperature?

Physics

1 answer:

TiliK225 [7]3 years ago

6 0

Thermal energy that flows between objects due to a difference in temperature is heat.

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It took 1500 Newton's of force to push a car 3 meters. How much work was done

denis23 [38]

Answer:

ow much work was done? W = F xD. IN X 2m = 2;. 2. A force of 15 newtons is ... 3. It took 50 joules to push a chair 5 meters across the floor. With what force was ... was done. How far was the rock lifted? W=FXD. D=1500 = 1.5m. Answer: :.5m ... A young man exerted a force of 9,000 newtons on a stalled car, but he was.

Explanation:

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3 years ago

To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp

motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

$\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0$

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that $f_{max}=\mu N$

Now, we find the acceleration:

$\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}$

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

$x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s$

8 0

3 years ago

Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by

OlgaM077 [116]

Explanation:

Area of ring $\ 2{\pi} a d a$

Charge of on ring $d q=-(\ 2{\pi} a d a)$

Charge on disk

$Q=-\left(\pi R^{2}\right)$

$\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}$