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s344n2d4d5 [400]
3 years ago
15

An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an

automobile of mass 890 kg traveling initially at a speed of 48.0 km/h in a distance equal to the diameter of a dime, which is 1.8 cm ?
Physics
1 answer:
Luba_88 [7]3 years ago
4 0

Answer:

F = 4399 KN

Explanation:

given,

mass of automobile = 890 kg

initial speed = 48 km/h

                     = 48 × 0.278 = 13.34 m/s

using equation of motion

v² = u² + 2 a × s

0 = 13.34² - 2 a ×0.018

a = \dfrac{13.34^2}{0.036}

a = 4943.21 m/s²

F   = m × a

F   = 890 × 4943.21

F   =  4399456.9 N

F = 4399 KN

hence, the  Net force is F = 4399 KN

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A ball of mass 8 kg falls from rest from a height of 100 m. Neglecting air resistance,calculate its kinetic energy after falling
astra-53 [7]
As an object falls from rest, its gravitational energy is converted to kinetic energy

G.P.E = K.E = mgh

K.E = (80 Kg)(9.8 m/s²)(30 m)

K.E. = 23,520 J
4 0
3 years ago
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A 87 arrow is fired from a bow whose string exerts an average force of 105 on the arrow over a distance of 75 .
timofeeve [1]

The solution would be like this for this specific problem:

 

V^2 = 2AS = 2FS/M

V = sqrt(2FS/M) = sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps

So the speed of the arrow as it leaves the bow is 42.5 mps.

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

6 0
3 years ago
The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. a typical axon carries a peak cu
Gemiola [76]

Answer:

6.66\cdot 10^{-12}T

Explanation:

The magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I is the current

r is the distance from the wire

In this problem we have

I=0.040 \mu A=4\cdot 10^{-8}A

r = 1.2 mm = 0.0012 m

So the magnetic field strength is

B=\frac{(4\pi \cdot 10^{-7} H/m)(4\cdot 10^{-8}A)}{2\pi (0.0012 m)}=6.66\cdot 10^{-12}T

5 0
3 years ago
Some trees have red leaves all year long. You know that plants with green leaves make their food for themselves. How do you thin
marshall27 [118]
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6 0
3 years ago
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A small block of mass m1 = 0.4 kg is placed on a long slab of mass m2 = 2.8 kg. Initially, the slab is stationary and the block
mel-nik [20]

Answer:

v₁ = 0.375 m / s ,   x = 0.335 m

Explanation:

Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.

We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.

initial moment. Right at the start of the movement

       p₀ = m v₀ + 0

final moment. Just when it comes to equilibrium

      p_{f} = (m + M) v₁

how the forces are internal

       p₀ =p_{f}

       m v₀ = (m + M) v₁

       v₁ = m /m+M    v₀

let's calculate

       v₁ = 0.4 /(0.4 + 2.8)  3

       v₁ = 0.375 m / s

 

Let's apply Newton's second law to the Block, to find the friction force

Y axis

       N - W = 0

       N = W

       N = m g

where m is the mass of the block

the friction force has the formula

      fr = μ N

      fr = μ m g

We apply Newton's second law to slab    

X axis

       fr = M a

where M is the mass of the slab

       μ m g = M a

       a = μ g m / M

let's calculate

       a = 0.15  9.8  0.4 / 2.8

       a = 0.21 m / s²

With kinematics we can find the position

       v²= v₀²+2 a x

as the slab is initially at rest, its initial velocity is zero

       v² = 2 a x

       x = v2 / 2a

let's calculate

        x = 0.375²/2 0.21

        x = 0.335 m

4 0
2 years ago
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