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dsp73
3 years ago
9

If Styrofoam has a specific heat of 1131 J/kg°C, how much energy would it take to change a 0.5 kg Styrofoam cooler from 15°C to

2°C?
Physics
1 answer:
Y_Kistochka [10]3 years ago
8 0

Answer:

.

Explanation:

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A ball is thrown vertically upwards with a velocity of 30m/s. Determine the maximum height reached
padilas [110]
The maximum height reached is 45.92 m

6 0
2 years ago
James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a
Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

4 0
3 years ago
The diagram shows the Earth rotating on it's axis. The two stars show different locations on the surface... How long does it tak
amid [387]
My guess would be about 10 years because stars are hot balls of light that are reflections from years ago so it would most likely take awhile
6 0
2 years ago
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Suppose we have a laser emitting a diffraction-limited beam (=632 nm) with a 2-mm diameter. How big a light spot from this lase
expeople1 [14]

Answer:

Explanation:

λ = wave length = 632 x 10⁻⁹

slit width a = 2 x 10⁻³ m

angular separation of central maxima

= 2 x λ /a

= 2 x 632 x 10⁻⁹ / 2 x 10⁻³

= 632 x 10⁻⁶ rad

width in m of light spot.

= 632 x 10⁻⁶  x 376000 km

= 237.632 km

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3 years ago
A microfarad capacitor is charged via a 12v battery. there is 1 kohms resistor in series with the capacitor. how much current is
olga2289 [7]
The answer completely depends on the number that belongs in the space before the word "microfarad".
7 0
3 years ago
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