Which of the following is a vector quantity (a)mass (b)speed (c)time (d)acceleration

9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o

Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

$m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.$

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

$F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).$

Now the force of friction is $F_s=\mu*N,$ where $N$ is the normal force and from the diagram it is $F_y=mg*cos(\theta).$

Thus $F_s=\mu*N=\mu*mg*cos(\theta).$

Therefore,

$F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).$

Substituting the value for $F_H,m,\mu, and \:\theta$ we get:

$F_{tot}= -38.63N.$

Now acceleration is simply

$a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.$

The negative sign indicates that the acceleration is directed up the incline.

Part B:

$d=\frac{1}{2} at^2$

Which can be rearranged to solve for t:

$t=\sqrt{\frac{2d}{a} }$

Substitute the value of $d=0.50m$ and $a=7.7m/s$ and we get:

$t=0.36s.$

which is our answer.

Notice that in using the formula to calculate time we used the positive value of $a$ , because for this formula absolute value is needed.

5 0

3 years ago

A mass of 250 N is on a piston of 2.0 m^2. What force is needed to lift this piston if the area of the second piston is 0.5 m^2?

prohojiy [21]

Answer:

Explanation:

The pressure at one end of the piston is equal to the pressure on the second piston.

Pressure = Force/Area

F1/A1 = F2/A2

Given

F1 = 250N

A1 = 2.0m²

A2 = 0.5m²

F2 = ?

Substituting the given values in the formula;

250/2 = F2/0.5

cross multiply

250*0.5 = 2F2

125 = 2F2

F2 = 125/2

F2 = 62.5N

Hence the force needed to lift this piston if the area of the second piston is 0.5 m^2 is 62.5N

8 0

3 years ago

The basic building block of life is

snow_tiger [21]

<span>The basic building block of life is molecules, more specifically macromolecules. There are four different macromolecules which could all be described as the building blocks of life, namely carbohydrates, proteins, nucleic acids and proteins. </span>

8 0

3 years ago

An object moving at a constant speed of 25 m/s is making a turn with a radius of curvature of 7 m (this is the radius of the "ki

prisoha [69]

Answer:

- 278.34 kg m/s^2

Explanation:

The rate of the change of momentum is the same as the force.

The force that an object feels when moviming in a circular motion is given by:

F = -mrω^2

Where ω is the angular speed and r is the radius of the circumference

Aditionally, the tangential velocity of the body is given as:

v = rω

The question tells us that

v = 25 m/s

r = 7m

mv = 78 kg m/s

Therefore:

m = (78 kg m/s) / (25 m/s) = 3.12 kg

ω = (25 m/s) / (7 m) = 3.57 (1/s)

Now, we can calculate the force or rate of change of momentum:

F = - (3.12 kg) (7 m)(3.57 (1/s))^2

F = - 278.34 kg m/s^2

4 0

3 years ago

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

soldi70 [24.7K]

$Given:\\T=29.46y\approx 9.29\cdot 10^8s\\M_S\approx2.0\cdot 10^{30}kg\\G=6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2} \\\\Find:\\R=?\\\\Solution:\\\\F_g= G\frac{mM_s}{R^2} \\\\F_c= \frac{mv^2}{R} \\\\F_g=F_c\\\\G\frac{mM_s}{R^2}=\frac{mv^2}{R} \Rightarrow G\frac{M_s}{R^2}=\frac{v^2}{R}\\\\v=\omega r\\\\G\frac{M_s}{R^2}= \frac{\omega^2R^2}{R}\Rightarrow G\frac{M_s}{R^2}=\omega^2R \\\\\omega= \frac{2 \pi }{T} \\\\G\frac{M_s}{R^3}= \frac{4 \pi ^2}{T^2}$

$GM_ST^2=4 \pi ^2R^3\Rightarrow R= \sqrt[3]{ \frac{GM_ST^2}{4 \pi ^2} }\\\\\\R= \sqrt[3]{ \frac{6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2}\cdot2.0\cdot 10^{30}kg( 9.29\cdot 10^8s)^2}{4\cdot 3.14^2} } \approx 1.42\cdot 10^{12}m$

3 0

3 years ago