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viktelen [127]
2 years ago
13

when striking, the pike, a predatory fish, can accelerate from rest to a speed of 4.0 m/sm/s in 0.15 ss

Physics
1 answer:
Alina [70]2 years ago
6 0

The distance covered will  be 0.301m.

firstly we will new to calculate acceleration (a).

we can do so using newtons first law of motion.

v=u + a × t

where

a is the acceleration

t is the time

from here we get

a=(v-u)/t

a=(4-0)/0.15

a=26.67m/s²

using newtons second law of motion

s=u.t+1/2at²

where s is distance travelled in t seconds.

we get,

s=0.301m

learn more about newtons law of motion here:

brainly.com/question/129361

#SPJ4

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<span>the state of matter of the medium i think
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a dog pulls on a pillow with a force of 8.4 N at an angle of 31 degrees above the horizontal. what is the x component of this fo
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The x component of the force is 8.4N * cos(31°) = 7.2N (2 s.f.)
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What forces affect a toy car as it moves down a ramp?
Amanda [17]

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gravity

Explanation:

3 0
3 years ago
Read 2 more answers
A bullet is shot vertically upward with an initial velocity of 128 ft/s. The bullet's height after t seconds is y(t) = 128t - 16
saul85 [17]

The height of the bullet when the velocity is zero is 256 ft.

<h3>Height of the bullet when the velocity is zero </h3>

The height of the bullet when the velocity is zero is determined by taking derivative of the function as shown below;

v = \frac{dy}{dt} = 128 -32t \\\\when \ v \ is \ zero\\\\v = 0\\\\128 - 32t = 0\\\\32t = 128\\\\t = \frac{128}{32} \\\\t = 4 \  s

The height of the bullet at this time is calculated as follows;

y(4) = 128(4) - 16(4)^2\\\\y(4) = 256 \ ft

Learn more about height of projectiles here: brainly.com/question/10008919

6 0
3 years ago
A rope of negligible mass passes over a uniform cylindrical pulley of 1.50kg massand 0.090m radius. The bearings of the pulley h
boyakko [2]

Answer:

Explanation:

mass of pulley, m3 = 1.5 kg

Radius of pulley, R = 0.09 m

mass of monkey, m2 = 4.5 kg

mass of banana bunch, m1 = 3 kg

Let a is teh acceleration ans T1 and T2 be the tension in the rope.

The moment of inertia of the pulley

I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²

According to Newton's second law

T1 - m1 g = m1 x a .... (1)

m2 g - T2 = m2 x a ..... (2)

(T2 - T1 ) x R = I x α    

where, α is the angular acceleration

α = a / R

(T2 - T1)R = 0.5 x m3 x R² x a / R

T2 - T1 = 0.5 x m3 x a ..... (3)  

from (1), (2) and (3)

a = \frac{m_{2}-m_{1}}{m_{1}+m_{2}+\frac{m_{3}}{2}}\times g

a = \frac{4.5-3}{3+4.5+0.75}\times 9.8

a = 1.78 m/s²

from equation (1)

T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N

from equation (2)

T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N

4 0
3 years ago
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