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3 years ago

Your grandmother in Ireland sends you her favorite cookie recipe. Her instructions say to bake the cookies at

Physics

2 answers:

Anna [14]3 years ago

7 0

Answer:

350 Degrees Fahrenheit

Explanation:

Lapatulllka [165]3 years ago

3 0

$F=\frac{9}{5}C+32$

$F=\frac{9}{5}190.5C+32$

$F=342.9+32$

$F=374.9$

$374.9F=190.5C$

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Un autocar que circula a 81 km/h frena uniformemente con una aceleración de -4,5 m/s2.

horsena [70]

Answer:

a) $\Delta x=56.25 m$

b) imagen adjunta

Explanation:

a) Primero debemos hacer la conversión de 81 km/h a m/s, esto es 22.5 m/s.

Ahora, usando la ecuacion cinemática, en un movimiento acelerado tenemos:

$v_{f}^{2}=v_{0}^{2}+2a \Delta x$

Queremos encontrar la posición hasta detenerse, osea vf = 0.

$\Delta x=\frac{-v_{0}^{2}}{2a}$

$\Delta x=\frac{-22.5^{2}}{-2*4.5}$

$\Delta x=56.25 m$

b) Para este caso el gráfico se encuentra adjunto.

Espero que te sirva de ayuda!

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8 0

3 years ago

A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force

Tamiku [17]

Answer:

(a) 91 kg (2 s.f.) (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

$s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}$

Subsequently,

$F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})$

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

$v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}$

According to Newton's First Law which states objects will remain at rest

or in uniform motion (moving at constant velocity) unless acted upon by

an external force. Hence, the block of ice by the end of the first 5

seconds, experiences no acceleration (a = 0) but travels with a constant

velocity of 4.4 $m \ s^{-1}$ .

$s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}$