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2 years ago

A book sitting on a table is moved horizontally. Describe the

Physics

1 answer:

Akimi4 [234]2 years ago

3 0

Frictional force and Applied force has same “magnitude” and “opposite” direction.

Option: B

<u>Explanation</u>:

When a book is moved horizontally by applying “force” on the book, the frictional force is opposed to the book by the table. Here, this “frictional force” is opposing the book has the same force what we applied on the book but this frictional force and the applied force are opposite in direction. Always the “frictional force” is opposite to the “applied force” which stops the object to move. For example, if a force applied leftward to the object the frictional force is acted on the right side of the object.

When two objects are in contact they experience a "frictional force". This "frictional force" acts opposite to the force applied on to move the object.

Formula for "frictional force" is $\mu\times N$

Where, $\mu$ is coefficient of friction and N is normal force.

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A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow

Ksivusya [100]

Answer:

The answer is 1.87nm/s.

Explanation:

The $110g/hr$ water loss must be replaced by $110g/hr$ of sap. 110g of sap corresponds to a volume of

$110g \div \dfrac{1040*10^3g}{1*10^6cm^3} = 106cm^3$

thus rate of sap replacement is

$106cm^3/hr = 106*10^{-6}m^3/3600s = 2.94*10^{-8}m^3/s$

The volume of sap in the vessel of length $x$ is

$V = Ax$ ,

where $A$ is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

$V = 2000Ax$

taking the derivative of both sides we get:

$\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}$

on the left-hand-side $\dfrac{dx}{dt}$ is the velocity $v$ of the sap, and on right-hand-side $\dfrac{dV}{dt} = 2.94*10^{-8}m^3/s$ ; therefore,

$2.94*10^{-8}m^3/s=2000Av$

and since the cross-sectional area is

$A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2$ ;

therefore,

$2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v$

solving for $v$ we get:

$v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}$

$\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}$

which is the upward speed of the sap in each vessel.

8 0

3 years ago

A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3

kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, $a=25.4\ m/s^2$

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

$v=u+at$

$v=25.4\times 3.39=86.10\ m/s$

Let x is the initial position of the rocket. Using third equation of kinematics as :

$v^2=u^2+2ax_o$

$x_o=\dfrac{v^2}{2a}$

$x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m$

Let $x_o$ is the position at the maximum height. Again using equation of motion as :

$v^2-u^2=2a(x-x_o)$

Now $a=-g$ and v and u will interchange

$u^2=2g(x-x_o)$

$x=x_o+\dfrac{u^2}{2g}$

$x=145.92+\dfrac{(86.10)^2}{2\times 9.8}$

x = 524.14 meters

Hence, this is the required solution.

5 0

3 years ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

2 years ago

A black squirrel is sitting on the ground. A red squirrel is sitting on a tree branch high above the black squirrel. Both squirr

labwork [276]

The black squirrel has zero kinetic energy (if it's not moving) and lower gravitational potential energy than the red squirrel or zero gravitational potential energy if the ground is assumed to be zero gravitational potential line.

4 0

2 years ago

Read 2 more answers

A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the

sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

$x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg$

Since the frictional force $f_k$ is equal to $f_k=\mu_k N=\mu_k mg$ , then we have that the acceleration is:

$a=\frac{\mu_k mg}{m}=\mu_k g$

Now, from the definition of acceleration we get:

$a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}$

And, as the final velocity is zero because the box gets to a stop, we have:

$t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}$

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

$t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s$

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

$v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}$

Finally, we calculate the displacement:

$x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm$

This means that the box moves 7.29cm backwards relative to the belt.

4 0

2 years ago