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3 years ago

Physics {deceleration}

Physics

1 answer:

seropon [69]3 years ago

7 0

Using kinematic equation, v^2 - u^2 = 2as. 5^2 - 3^2 = 2a x 16. a = 0.5m/s^2. So particle will deaccelerate at 0.5m/s^2. ( v = final velocity, u= initial velocity, a= acceleration, s= displacement.)

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The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of t

zysi [14]

Answer:

Explanation:

Speed of electrical nerve signal = 33 m /s

Distance travelled = 1.3 m

time taken = distance / speed

= 1.3 / 33

= .039 s

= 39 ms ( millisecond ) .

3 0

3 years ago

We push a 38.4-kg box across the floor at constant velocity. If we are pushing a horizontal force of 238 N, find the coefficient

zmey [24]

Answer: 0.62

Explanation:

Coefficient of friction is defined as the ratio of the moving force (Fm) acting on a body to the normal reaction (R).

Note that the normal reaction acts vertically on the object and is equal to the objects weight (W) i.e W=R

Since W = mg, W = 38.4 ×10

W= 384N =R

Normal reaction = 384N

The horizontal force acting on the body will be the moving force which is 238N

Coefficient of friction = Fm/R

Coefficient of friction = 238/384

Coefficient of friction = 0.62

Therefore, coefficient of kinetic friction between the box and the floor is 0.62

5 0

3 years ago

Find the y-component of this

Alborosie

Answer:

-0.0789 m

Explanation:

Recall that the y-component comes associated with the sin(18.4) through the following trigonometric relationship:

y = 0.250 sin(-18.4) ≈ -0.0789 m

Notice it is negative since it is below the x-axis.

4 0

3 years ago

A certain wave motion has a frequency of 10 cycles / s and a wavelength of 3 m. so its propagation speed is

kodGreya [7K]

3.5m is ur answer ask for more questions anytime

5 0

3 years ago

A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is

goblinko [34]

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, $v_i$ = 1.4 m/s

final velocity of the child, $v_f$ = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

total mass, m = 48 kg

frictional force, $f_k$ = 41 N

The work done by the child is calculated as;

$\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech} + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96 \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J$

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

5 0

3 years ago