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2 years ago

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Everyone i need help so much

2 answers:

Veronika [31]2 years ago

3 0

Answer:

b

Explanation:

katrin [286]2 years ago

3 0

Answer:

b I got it right on this quiz I did one time

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A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as

Taya2010 [7]

Answer:

$w = - 508.53$ joules

$q = - 3091.47$ joules

Explanation:

Let us convert the time in hours into seconds

$0.010* 3600\\= 36$

Change in internal energy

$\delta E = p * \delta t$

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

$\delta E = - 100 * 36\\$

$\delta E = - 3600$ Joules

Amount of work done by the system

$w = - P * \delta V$

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

$w = - 1 * ( 5.92 -0.90)\\$

$w = -5.02$ liter-atmospheres

Work done in Joules

$- 5.02 * 101.3\\$ $= 508.53$ Joules

$q = \delta E - w\\$

Substituting the given values we get -

$q = - 3600 - (-508.53)\\q = - 3091.47$

Thus

$w = - 508.53$ joules

$q = - 3091.47$ joules

7 0

3 years ago

A cyclist accelerates from a velocity of 10 miles/hour east until reaching a velocity of 20 miles/hour east in 5 seconds. What w

Sliva [168]

Answer:

$a = 0.894\ m/s^2$

Explanation:

<u>Motion with Constant Acceleration</u>

A body moves with constant acceleration when the speed changes uniformly in time. The equation used to find the final speed vf is

$v_f=v_o+at$

Where vo is the initial speed, a is the acceleration, and t is the time.

The cyclist has an initial speed of vo=10 miles/hour and ends up at vf=20 miles/hour in t=5 seconds.

Both speeds are given in miles/hour and we must convert it to m/s:

1 mile/hour = 0.44704 m/s

10 mile/hour = 4.47 m/s

20 mile/hour = 8.94 m/s

The acceleration is calculated by solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

$\displaystyle a=\frac{8.94-4.47}{5}$

$a = 0.894\ m/s^2$

3 0

2 years ago

Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th

timama [110]

Answer:

Explanation:

Given

length of window $h=2.9\ m$

time Frame for which rock can be seen is $\Delta t=0.134\ s$

Suppose h is height above which rock is dropped

Time taken to cover $h+2.9 is t_1$

so using equation of motion

$y=ut+\frac{1}{2}at^2$

where y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h is

$h=0+0.5\times g\times (t_2)^2---2$

Subtract 1 and 2 we get

$2.9=0.5g(t_1^2-t_2^2)$

$5.8=g(t_1+t_2)(t_1-t_2))$

and from equation $t_1-t_2=0.134\ s$

so $t_1+t_2=\frac{5.8}{9.8\times 0.134}$

$t_1+t_2=4.416\ s$

and $t_1=t_2+\Delta t$

so $t_2+\Delta t+t_2=4.416$

$2t_2+0.134=4.416$

$t_2=0.5\times 4.282$

$t_2=2.141\ s$

substitute the value of $t_2$ in equation 2

$h=0.5\times 9.8\times (2.141)^2$

$h=22.46\ m$

8 0

3 years ago

physics A river flows at a speed vr = 5.37 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shorelin

vova2212 [387]

Answer: Vb is the vector (-5.37m/s, 8.59 m/s), with a module 10.13m/s

then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9

Explanation:

We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.

We have Vr = (5.37m/s, 0m/s)

Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).

The velocity of the boat Vb will be:

Vb = (-c*sin(32). c*cos(32))

Then we should have that:

5.37 m/s - c*sin(32) = 0

c = (5.37/sin(32))m/s = 10.13 m/s

the velocity in the y-axis is:

10.13m/s*cos(32) = 8.59 m/s

So Vb = (-5.37m/s, 8.59 m/s)

the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.

7 0

3 years ago

(6) Module 4 Study Guide

velikii [3]

We have the meats Arby’s we beat them kids

3 0

2 years ago