How to figure out the question?

A 20 kg wagon is pulled along the level ground by a rope inclined at 30 degree above the horizontal. A friction force of 30 N op

Elan Coil [88]

(a) 34.6 N

To solve the problem, we have to analyze the forces acting along the horizontal direction.

We have:

- Forward: the component of the pull parallel to the ground, which is

$F cos \theta$

where

F is the magnitude of the pull

$\theta=30^{\circ}$ is the angle

- Backward: the force of friction, which is

$F_f = 30 N$

So, the equation of motion is

$F cos \theta - F_f = ma$

where

m = 20 kg is the mass of the wagon

a is the acceleration

In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

$F cos \theta - F_f = 0$

Therefore, we can find the pulling force:

$F=\frac{F_f}{cos \theta}=\frac{30}{cos 30}=34.6 N$

(b) 43.9 N

In this case, the acceleration is

$a=0.40 m/s^2$

So, the equation of motion in this case is

$F cos \theta - F_f = ma$

So this time we have to take into account the term (ma).

Using the same data as before:

m = 20 kg

$\theta=30^{\circ}$

$F_f = 30 N$

We find the new magnitude of F:

$F=\frac{ma+F_f}{cos \theta}=\frac{(20)(0.40)+30}{cos 30}=43.9 N$

6 0

3 years ago

A sample of gas has an initial volume of 4.5 L at a pressure of 754 mmHg . Part A If the volume of the gas is increased to 8.5 L

Firdavs [7]

Answer:

The pressure will be of 399.17 mmHg.

Explanation:

p1= 754 mmHg

V1= 4.5 L

p2= ?

V2= 8.5 L

p1*V1 = p2*V2

p2= (p1*V1)/V2

p2= 399.17 mmHg

6 0

3 years ago

What do you think we can learn about Earth's history by looking at the Grand Canyon?

viktelen [127]

We can look at all the ages of the earth since it’s a big crack is reveals many layers of the earth and we can know about chemicals and metals that were in earth and diffrent times

5 0

3 years ago

A+10 u charge and a -10 4C (1 HC - 106 C), at a distance of 0.3 m,

Marina CMI [18]

Answer:

B. Attract each other with a force of 10 newtons.

Explanation:

Statement is incorrectly written. The correct form is: A $+10\,\mu C$ charge and a $-10\,\mu C$ at a distance of 0.3 meters.

The two particles have charges opposite to each other, so they attract each other due to electrostatic force, described by Coulomb's Law, whose formula is described below:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$ (1)

Where:

$F$ - Electrostatic force, in newtons.

$\kappa$ - Electrostatic constant, in newton-square meters per square coulomb.

$|q_{A}|,|q_{B}|$ - Magnitudes of electric charges, in coulombs.

$r$ - Distance between charges, in meters.

If we know that $\kappa = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $|q_{A}| = |q_{B}| = 10\times 10^{-6}\,C$ and $r = 0.3\,m$ , then the magnitude of the electrostatic force is:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$

$F = 9.987\,N$

In consequence, correct answer is B.

4 0

3 years ago

a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t

lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0

2 years ago