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son4ous [18]
4 years ago
15

Define work done against friction(helpp)​

Physics
1 answer:
77julia77 [94]4 years ago
7 0

Answer:

The work done against friction is the work done on an object that overcomes this frictional force allowing the object to move - it doesn't include any extra work used to accelerate an object for example - only the work used to beat the frictional force.

Explanation:

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Two particles, one with charge -6.29 × 10^-6 C and one with charge 5.23 × 10^-6 C, are 0.0359 meters apart. What is the magnitud
navik [9.2K]

Answer:

Force, F = −229.72 N

Explanation:

Given that,

First charge particle, q_1=-6.29\times 10^{-6}\ C

Second charged particle, q_2=5.23\times 10^{-6}\ C

Distance between charges, d = 0.0359 m

The electric force between the two charged particles is given by :

F=k\dfrac{q_1q_2}{d^2}

F=9\times 10^9\times \dfrac{-6.29\times 10^{-6}\times 5.23\times 10^{-6}}{(0.0359)^2}

F = −229.72 N

So, the magnitude of force that one particle exerts on the other is 229.72 N. Hence, this is the required solution.

4 0
3 years ago
What is the total displacement of the car after 5 h? Question 4 options: 0 km 15 km 20 km 40km.
Likurg_2 [28]

Displacement can be defined as the overall change in the position of the object. The displacement of the car in 5 hrs is 0 km.

Displacement:

It is a vector quantity that is defined as the overall change in the position of the object.

The formula for displacement is:

s = s_f-s_i

Where,

s - total displacement

s_f - final position

s_i- initial position

In the given graph:

  • Car displaces 15 km in 2 hrs
  • Car displaces 20 km in 4 hrs.
  • The car moves back to the initial position in 5 hrs.

Therefore, the displacement of the car in 5 hrs is 0 km.

Learn more about Displacement:  

brainly.com/question/10919017

3 0
3 years ago
A student measures the length of the shadow of the Washington Monument to be 620 ft. If the Washington Monument is 555 ft tall,
Nuetrik [128]

Answer:

The angle of elevation of the Sun = 41.8°

Explanation:

Let the angle of elevation of the sun be θ

Using trigonometric relations,

Tan θ = opp/adj

opp = the height of the Washington Tower = 555 ft

adj = the length of the shadow = 620 ft

Tan θ = (555/620) = 0.8952

θ = Tan⁻¹ (0.8952) = 41.8°

Hence, the angle of elevation of the Sun = 41.8°

8 0
3 years ago
Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.
Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

f=\frac{R}{2}

For this mirror, R = 20 cm, so its focal length is

f=\frac{20}{2}=+10 cm (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm

b)

We can solve this part by using the magnification equation:

M=-\frac{y'}{y}=\frac{q}{p}

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm

and the negative sign means that the image is inverted.

#LearnwithBrainly

6 0
3 years ago
The Earth's average density a. can be deduced from its volume and gravitational pull. b. is the same as rocks on the surface. c.
tester [92]

Answer:

option A

Explanation:

Earth density can be deduced with the help of gravitational pull and the volume.

gravitational pull of the earth is equal to= \dfrac{GM}{R^2}........(1)

We know, the volume of the earth, with the help of that we can calculate radius of the earth.

We know gravitational pull on earth = 9.8 m/s²

From equation (1) we can calculate the mass.

So, Density = mass/ Volume.

From the above expression we can calculate the density of earth.

Hence, the correct answer is option A

3 0
3 years ago
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