Select the correct answer?

Heidi (39 kg) is walking her tiny chihuahua, Chaxi (5.60 kg), on the sidewalk. To encourage Chaxi along, Heidi pulls forward wit

Ivan

Answer:

The correct reaction force in response to Heidi's action force is:

c. The friction is equal to 660 N since the beam is not accelerating.

Explanation:

Heidi's action force does not affect the beam. Since friction resists the sliding or rolling of one solid object over another, there is no friction acting on the beam, in this respect. The reaction force is what makes the dog to move because it acts on it. According to Newton's Third Law of Motion, forces always come in action-reaction pairs. This Third Law states that for every action force, there is an equal and opposite reaction force. This means that the dog exerts some force on Heidi, as he pulls it "forward with a force of 9.55 N."

4 0

2 years ago

Based on this passage, what is campylobacter?

alexdok [17]

I believe the correct answer from the choices listed above is option 4. Base from the passage given above, it is very clear that the Campylobacter is a bacteria used to test in <span>identifying drug-resistant bacteria. Hope this answers the question. Have a nice day.</span>

4 0

3 years ago

Read 2 more answers

When a container decreases in size, what will happen to an amount of gas in the container??

alexgriva [62]

The gas would also decrease in size since the container lost gas to decrease the size of the container.

7 0

3 years ago

Which calculates the intensity of an electric field at a point where a 0.50 C charge experiences a force of 20. N?

tester [92]

Answer: $40\ N/C$

Explanation:

Given

Magnitude of charge is $q=0.5\ C$

Force experienced is $F=20\ N$

Electric field intensity is the electrostatic force per unit charge

$\therefore E=\dfrac{F}{q}\\\\\Rightarrow E=\dfrac{20}{0.50}\\\\\Rightarrow E=40\ N/C$

Thus, the electric field intensity is $40\ N/C$

6 0

3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago