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mars1129 [50]
3 years ago
10

A 50.0 kg object rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and

the coefficient of kinetic friction is 0.200. What is the friction force on the object if a horizontal 140 N push is applied to it?
Physics
1 answer:
MariettaO [177]3 years ago
8 0

Answer:

f=140\ N

Explanation:

Given:

  • mass of the object on a horizontal surface, m=50\ kg
  • coefficient of static friction, \mu_s=0.3
  • coefficient of kinetic friction, \mu_k=0.2
  • horizontal force on the object, F=140\ N

<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>

F_s=\mu_s.N

where:

N= normal force of reaction acting on the body= weight of the body

F_s=0.3\times (50\times 9.8)

F_s=147\ N

As we know that the frictional force acting on the body is always in the opposite direction:

So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.

so, the frictional force will be:

f=140\ N

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The sound wave will have traveled 2565 m  farther in water than in air.

Answer:

Explanation:

It is known that distance covered by any object is directly proportional to the velocity of the object and the time taken to cover that distance.

Distance = Velocity × Time.

So if time is kept constant, then the distance covered by a wave can vary depending on the velocity of the wave.

As we can see in the present case, the velocity of sound wave in air is 343 m/s. So in 2.25 s, the sound wave will be able to cover the distance as shown below.

Distance = 343 × 2.25 =771.75 m

And for the sound wave travelling in fresh water, the velocity is given as 1483 m/s. So in a time interval of 2.25 s, the distance can be determined as the product of velocity and time.

Distance = 1483×2.25=3337 m.

Since, the velocity of sound wave travelling in fresh water is greater than the sound wave travelling in air, the distance traveled by sound wave in fresh water will be greater.

Difference in distance covered in water and air = 3337-772 m = 2565 m

So the sound wave will have traveled 2565 m  farther in water than in air.

5 0
3 years ago
The resistance between 2 points in an electrical circuit is 1.1 Ω. What additional resistance
valentinak56 [21]

Answer:

the new resister is 11 ohms.

Explanation:

Set it  up like this.

1/x + 1/1.1 = 1                    Subtract 1/1.1 from both sides

1/x = 1 - 1/1.1

1 - 1/1.1 = 1/11

1/x = 1/11                          Cross multiply

11 = x

If 1/11 bothers you, you could do it it another way.

1 - 1/1.1 = (1.1 - 1 ) / 1.1 = 0.1 / 1.1  Multiply top and bottom by 10

0.1*10/(1.1 * 10 ) = 1 / 11

5 0
2 years ago
A chocolate chip cookie is an example of a (2 points) a homogeneous mixture b heterogeneous mixture c suspension d colloid
kumpel [21]

Answer:

I think it is heterogeneous mixture. have a good day

8 0
2 years ago
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To practice Problem-Solving Strategy 8.1 for circular-motion problems. A cyclist competes in a one-lap race around a flat, circu
Anna007 [38]

Answer:

Explanation:

distance travelled

s = 2πR

= 2 X 3.14 X 140

= 880 m

final velocity = v

initial velocity = u

distance travelled = s

time = 60 s

s = ut + 1/2 at²

880 = .5 x a x 60²

a = .244 m/s²

final velocity v = at

= .244 x 60

= 14.66

centripetal acceleration at final moment

v² /R

14.66 X 14.66 / 140

= 1.53 m/s⁻²

1.53 m/s²

this is centripetal acceleration which acts towards the centre.

tangential acceleration calculated a _t = .244

redial acceleration ( centripetal ) = 1.53

Resultant acceleration

R²= 1.53² + .244 ²

R = 1.55 m/s²

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5 0
3 years ago
A crate of oranges with a total mass of 6.7 kg is being pulled across a skating rink (frictionless) with a rope that makes an an
Mama L [17]

Answer:

36.74 N

Explanation:

Given that:

A crate of oranges with a total mass  (m) = 6.7 kg

angle θ = 0.7 rad

angle θ = 0.7 * \frac{180}{\pi}

angle θ = 40°

acceleration = 4.2 m/s²

Given that:

T cosθ  = ma

T cos 40° = 6.7 × 4.2

T = \frac { 6.7 * 4.2}{cos \ 40}

T = \frac { 28.14}{0.7660}

T = 36.74 N

Thus, the tension in the rope = 36.74 N

4 0
3 years ago
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