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ruslelena [56]
2 years ago
7

While you are studying for an upcoming physics exam, a lightning storm is brewing outside your window. Suddenly, you see a tree

across the street struck by lightning. There is a loud sound and you see smoke rising from various parts of the tree. You stop studying for your exam and do online research on lightning, trees, and sap. You find that a typical lightning bolt represents a potential difference of 1.00 ✕ 108 V between the cloud and the ground and that it can transfer a typical charge of 50.0 C between the cloud and the ground. Some models show that when a tree is hit by lightning, perhaps 2.00% of the energy in the lightning bolt can be delivered to the sap, causing it to boil. Model the sap as water initially at 30.0°C. If all of the sap in the tree is vaporized to gaseous sap at 100°C, determine how much sap (in kg) there is in the tree. (Use any necessary values found in this table or this table.)
Physics
1 answer:
vovikov84 [41]2 years ago
7 0
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A 4 kg box is on a frictionless 35° slope and is connected via a massless string over a massless, frictionless pulley to a hangi
Anarel [89]

Answer:

(a) 19.62 N

(b) Box moves down the slope

(c) 24.43 N

Explanation:

(a)  

2 Kg box  causes tension

T=mgwhere m is mass, g is gravitational force taken as 9.81T=2*9.81 =19.62 N  (b)  Block mass of 4 Kg  [tex]T'-mg sin \theta=0 hence T'=mg sin \theta where m is mass and g is gravitational force  

T'=4*9.81 sin 35= 22.5071 N  

Since T' is greater than mg sin\theta , then the box moves down the slope  

(c)  

Acceleration a= \frac {forward   force-backward   force}{Total mass}= \frac {mg sin \theta -mg}{m1 + m2}  

a= \frac {22.51-19.62}{2+4}=0.48

When moving, the box will exert force T"= mgsin \theta + ma  

T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N

7 0
2 years ago
Read 2 more answers
A small mass m is tied to a string of length L and is whirled in vertical circular motion. The speed of the mass v is such that
adell [148]

Answer:

(mv^2/R)/(mg)=1/2

v^2=R/2g

7 0
2 years ago
for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha
VLD [36.1K]

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

R=\dfrac{u^2\ \sin2\theta}{g}

Here,

u is the speed of launch of projectile

Here, \theta=30^{\circ}

We need to find the other launch angle when the projectile have the same range, such that,

\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}

\sin(60)=\sin2\alpha

\dfrac{\sqrt3}{2}=\sin2\alpha

\alpha =30^{\circ}

So, the other angle is 30 degrees. Hence, this is the required solution.

3 0
2 years ago
A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the
hoa [83]

Answer:

Since the spring mass system will execute simple harmonic motion the position as a function of time can be written asx(t)=Asin(\omega t+\phi)

'A' is the amplitude = 6 inches (given)

\omega =\sqrt{\frac{k}{m}} is the natural frequency of the system

At equilibrium we have

mg=kx\\\\k=\frac{mg}{x}

Applying values we get

k=40 lb/ft

thus natural frequency equals

\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}

Thus the equation of motion becomes

x(t)=6sin(8t+\phi)

At time t=0 since mass is at it's maximum position thus we have

A=Asin(\omega t+\phi)\\\\\therefore sin(\omega\times 0+\phi)=1\\\\\phi=\frac{\pi}{2}\\\\\therefore x(t)=Asin(\omega t+\frac{\pi}{2})

Thus the position of mass at the given times is as follows

1) at \frac{\pi}{12} x(t)=5.99inches

2) at \frac{\pi}{8} x(t)=5.9909inches

3) at \frac{\pi}{6} x(t)=5.98397inches

4) at \frac{\pi}{4} x(t)=5.9639inches

5) at \frac{9\pi}{32} x(t)=5.954inches

4 0
1 year ago
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Sergio039 [100]

Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 4=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4\times 2}{9.81}}\\\Rightarrow t=0.903\ s

s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2

The acceleration of Superman would be -963.93 m/s² from Lois' perspective

6 0
1 year ago
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