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Amiraneli [1.4K]
1 year ago
7

Using a horizontal force of 200 n, we intend to move a wooden cabinet across a floor at a constant velocity. What is the frictio

n force that will be exerted on the cabinet?.
Physics
1 answer:
grin007 [14]1 year ago
5 0

By using the second law of Newton, the frictional force is 200N.

We need to know about the second law of Newton (force) to solve this problem. The total force applied an object is proportional to the mass of object and acceleration. It can be defined as

∑F = m . a

where F is force, m is mass and a is acceleration.

From the question above, we know that

F1 = 200N

v = constant therefore (a = 0 m/s²)

By using second law of Newton, we get

∑F = m . a

F1 - Ffriction = m . 0

200 - Ffriction = 0

Ffriction = 200 N

Hence, the frictional force is 200N.

Find more on force at: brainly.com/question/25239010

#SPJ4

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Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl
lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

   m g sin\theta - F_{rope} = ma

 velocity is constant, a = 0

   m g sin\theta - F_{rope} =0

   F_{rope} = m g sin\theta

   F_{rope} = 70.1\times 9.8\times sin 8.6^0

  F_{rope}= 102.73\ N

b) now, when acceleration, a = 0.135 m/s²

   F_{rope}-m g sin\theta = ma

 velocity is constant, a = 0.135 m/s₂

   F_{rope} = m g sin\theta+ma

   F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135

  F_{rope}= 112.19\ N

7 0
3 years ago
A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.550 m/s. The to
Arisa [49]

Answer: 0.5 m/s

Explanation:

Given

Speed of the sled, v = 0.55 m/s

Total mass, m = 96.5 kg

Mass of the rock, m1 = 0.3 kg

Speed of the rock, v1 = 17.5 m/s

To solve this, we would use the law of conservation of momentum

Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns

When the man throws the rock forward

rock:

m1 = 0.300 kg

V1 = 17.5 m/s, in the same direction of the sled with the man

m2 = 96.5 kg - 0.300 kg = 96.2 kg

v2 = ?

Law of conservation of momentum states that the momentum is equal before and after the throw.

momentum before throw = momentum after throw

53.08 = 0.300 * 17.5 + 96.2 * v2

53.08 = 5.25 + 96.2 * v2

v2 = [53.08 - 5.25 ] / 96.2

v2 = 47.83 / 96.2

v2 = 0.497 ~= 0.50 m/s

3 0
3 years ago
An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance
marshall27 [118]

Answer:

E = \frac{\lambda}{2\pi \epsilon_0 r}

Explanation:

Let the linear charge density of the charged wire is given as

\frac{q}{L} = \lambda

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

\int E. dA = \frac{q}{\epsilon_0}

here we have

E \int dA = \frac{\lambda L}{\epsilon_0}

E. 2\pi r L = \frac{\lambda L}{\epsilon_0}

so we have

E = \frac{\lambda}{2\pi \epsilon_0 r}

4 0
3 years ago
What is the change in momentum of a 50-kg rock that falls freely for 3 seconds?
Elan Coil [88]

Answer:

1470kgm/s

Explanation:

Given parameters:

Mass of the rock = 50kg

Time taken for the free fall  = 3s

Unknown:

Change in momentum = ?

Solution:

The change in momentum will be difference between the ending momentum and finishing momentum.

  Momentum is the product of mass and velocity

       Momentum  = mass x velocity

Initial momentum  = 0, the velocity is 0

Final momentum = mass x final velocity

      let us find the final velocity;

                V = U + gt

V is the final velocity

U is the initial velocity

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

t is the time

                 V  = 0 + 9.8x3 = 29.4m/s

So;

 Change in momentum  = 50 x 29,4  = 1470kgm/s

6 0
3 years ago
A solid 200-g block of lead and a solid 200-g block of copper are completely submerged in an aquarium filled with water. Each bl
finlep [7]

Answer:

B. The buoyant force on the copper block is greater than the buoyant force on the lead block.

Explanation:

Given;

mass of lead block, m₁ = 200 g = 0.2 kg

mass of copper block, m₂ = 200 g = 0.2 kg

density of water, ρ = 1 g/cm³

density of lead block, ρ₁ = 11.34 g/cm³

density of copper block, ρ₂ = 8.96 g/cm³

The buoyant force on each block is calculated as;

F_B = mg(\frac{density \ of \ fluid}{density \ of \ object} )

The buoyant force of lead block;

F_{lead} = 0.2*9.8(\frac{1}{11.34} )\\\\F_{lead} = 0.173  \ N

The buoyant force of copper block

F_{copper} = 0.2*9.8(\frac{1}{8.96})\\\\F_{copper} = 0.219  \ N

Therefore, the buoyant force on the copper block is greater than the buoyant force on the lead block

3 0
3 years ago
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