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denis-greek [22]
3 years ago
9

A 4 kg block has a coefficient of friction of 0.3 between itself and the surface

Physics
1 answer:
Deffense [45]3 years ago
4 0

Answer:

C - 11.76 N (Newtons)

Explanation:

2 steps are needed,

First find the Force being applied by the box on the surface.

F = M(a) [Force equals: mass X Acceleration (of gravity)]

You get 39.2 N of Force.

Plug in the remaining values into the equation for Frictional Force

F(f) = μ(F) [Coefficient of Friction X Force applied by the Box]

So

F(f) = 0.3(39.2) = 11.76 N (of Frictional Force)

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A dog (mass 12.3 kg) sits on a sled (mass 5.1 kg) that is being pushed with a force of 33 N on a flat snowy surface. If the acce
GarryVolchara [31]
First, you need to calculate the resultant force:
R = m · a
   = (12.3 + 5.1) · 1.5
   = 26.1 N

Then, you can calculate the force of friction:
R = F - Fₐ
Fₐ = F - R
    = 33 - 26.1
    = 6.9 N

Now, we know that:
Fₐ = μ·m·g
Therefore we can solve for μ:

μ = <span>Fₐ/mg
   = 6.9 / (17.4 · 9.8)
   = 0.40 

The coefficient of dynamic friction is </span>μ = <span>0.40
</span>
4 0
3 years ago
A 40.0-$kg$ body is moving in the direction of the positive x axis with a speed of 238 $m/s$ when, owing to an internal explosio
Julli [10]

Answer:

v_3 = 400\ m/s

Explanation:

given,

mass of the body = 40 Kg

speed in x-axis = 238 m/s

mass break into three part

m₁ = 7 kg

v₁ = 356 m/s (along the positive y axis)

m₂ = 4.5 kg

v₂ = 357 m/s(along the negative x axis)

m₃ = 40 - (7 + 4.5) = 28.5 Kg

v₃ = ?

using conservation of momentum

MV = m₁v₁ + m₂v₂ + m₃v₃

(40)(238) \hat{i} = (7)(356) \hat{j} - (4.5)(357) \hat {i} + 28.5 v_3

(9520) \hat{i} = 2492 \hat{j} - 1606.5\hat {i} + 28.5 v_3

11126.5 \hat{i} - 2492 \hat{j} = 28.5 v_3

v_3 =390.40 \hat{i} - 87.44 \hat{j}

v_3 = \sqrt{390.40^2 + 87.44^2}

v_3 = 400\ m/s

4 0
3 years ago
'can you outrun a tornado ??
jekas [21]

Answer:

No, because the tornado goes faster than a car at 100 m per hour

Explanation:

7 0
3 years ago
Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a const
Dafna1 [17]

Answer:

\frac{dA}{dt} = 188.5 m^2/s

Explanation:

As we know that area of the circle at any instant of time is given as

A = \pi r^2

now in order to find the rate of change in area we will have

\frac{dA}{dt} = 2\pi r\frac{dr}{dt}

here we know that

rate of change of radius is given as

\frac{dr}{dt}= 1 m/s

radius of the circle is given as

r = 30 m

now we have

\frac{dA}{dt} = 2\pi (30)(1)

\frac{dA}{dt} = 60\pi

\frac{dA}{dt} = 188.5 m^2/s

8 0
4 years ago
Can anyone answer this its a science question NO LINKS!!!
N76 [4]

Answer: heat to mechanical to electrical

Explanation:

5 0
3 years ago
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