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lara31 [8.8K]
2 years ago
6

This rock, known as balance rock, sits on a thin spike of rock in a canyon in idaho. explain the forces that keep the rock balan

ced on its tiny pedestal.
Physics
1 answer:
Evgen [1.6K]2 years ago
7 0

Answer:

Action force and Reaction force

Explanation:

The action force which is the balanced rock pushing down due to gravity and the reaction force pushing the equal amount of force. These two things are stated in Newtons third law, where he states that "Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first." Also, he states that "all forces acts in pairs," meaning that every force exerted, there is an opposite force on the first.

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A camera gives a proper exposure when set to a shutter speed of 1/250 s at f-number F8.0. The photographer wants to change the s
Oksana_A [137]

Answer:

F4.0

Explanation:

To obtain a shutter speed of 1/1000 s to avoid any blur motion the f-number should be changed to F4.0 because the light intensity goes up by a factor of 2 when the f-number is decreased by the square root of 2.

5 0
2 years ago
A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte
Vitek1552 [10]

Answer:

the internal energy of the gas is 433089.52 J

Explanation:

let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.

the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

     = 3/2×(5.3)×(8.31451)×(24 + 273)

     = 433089.52 J

Therefore, the internal energy of this gas is 433089.52 J.

5 0
2 years ago
A BMX bicycle rider takes off from a ramp at a point 2.4 m above the ground. The ramp is angled at 40 degrees from the horizonta
adoni [48]

Answer:

The BMX lands 5.4 m from the end of the ramp.

Explanation:

Hi there!

The position of the BMX is given by the position vector "r":

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t

x0 = initial horizontal position

v0 = initial velocity

α = jumping angle

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

Please, see the attached graphic for a better understanding of the situation. At final time, when the bicycle reaches the ground, the vector position will be "r final" (see figure). The y-component of the vector "r final" is - 2.4 m (placing the origin of the frame of reference at the jumping point). With that information, we can use the equation of the y-component of the vector "r" (see above) to calculate the time of flight. With that time, we can then obtain the x-component (rx in the figure) of the vector "r final". Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

-2.4 m = 0 m + 5.9 m/s · t · sin 40° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 5.9 m/s · t · sin 40° + 2.4 m

Solving the quadratic equation:

t = 1.2 s

Now, we can calculate the x-component of the vector "r final" that is the horizontal distance traveled by the bicycle:

x = x0 + v0 · t · cos α

x = 0 m + 5.9 m/s · 1.2 s · cos 40°

x = 5.4 m

The BMX lands 5.4 m from the end of the ramp.

Have a nice day!

8 0
2 years ago
A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 270 ω at room temperature, what is
vova2212 [387]

solution:

consider the following data\\
length of slicon bar with circular cross section is 4cm or 0.04m\\
at room temperature resistance of the slicon bar is 270\Omega \\
represent the resistance in mathematical from\\
r=p\frac{1}{A}---1\\
where r is resistance and l is the length \\
A is cross sectional area\\
it is clear that resistivity of the silicon meterial is 6.4\times^2 \Omega.m\\
substitute 6.4\times10^2 for p,270\Omega for R and 0.04m for l i equation (1).\\270=(604\times10^2)\frac{0.04}{A}\\
rewrite the equation\\
a=(6.4\times10^2)\frac{(0.04)}{270}\\
=0.9481m^2\\
write the formula for the circular cross sectional area of silicon bar.\\
A=\pi r^2\\
substitute 0.9481 for A in the above equation\\
\pi r^2=0.9481
r^2=\frac{0.9481}{3.14},since \pi =3.14\\
0.30194\\
further simplified\\
r^2=0.30194\\
\sqrt{0.30194}\\
\cong 0.1509m\\
\cong 150.1mm

7 0
2 years ago
Your friend states in a report that the average time required to circle a 1.5-mi track was 65.414 s. This was measured by timing
Aleks04 [339]

The time per lap was calculated by measuring the time for seven laps and dividing the total time by seven.

total time 65.414 s \times 7 =457.898 s.

It is given that the precision is of 0.1 s. it means it is correct upto 1 place beyond decimal.

So, the actual value could vary from 457.800 s to 457.899 s i.e. the time per lap could be 65.400 s to 65.414 s

It means measured time 65.414 s has maximum error of 0.021%. Hence, the measured value is quite precise.

5 0
2 years ago
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