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viktelen [127]
1 year 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]1 year 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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A golf ball stays on the tee until the golf club hits it. Which of the following principles best describes why this occurs? Grou
givi [52]

Answer:

Newton's First Law of Motion.

Explanation:

Newton's first law of motion states that an object continues to stay in its state of rest, or of uniform motion, until acted upon by an external force.

So in the case of the golf ball here, the ball stays in its state of rest, on the tee, until the golf club hits it, i.e. , applies an external force on it.

Hence we can say that Newton's First Law of Motion is the principle which is most suitable for explaining this phenomenon.

7 0
3 years ago
A train is pulling four train cars and each car has a mass of 40,000 kg. The train is accelerating at 1.1 m/s^2. What is the for
IgorLugansk [536]

Answer:

176,000 N

Explanation:

Newton's second law:

∑F = ma

F = (4 × 40,000 kg) (1.1 m/s²)

F = 176,000 N

8 0
3 years ago
Read 2 more answers
Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

5 0
2 years ago
A man pushes his lawnmower with a velocity of +0.75 m/s relative to the ground. A girl rides by on her bike with a velocity of +
SashulF [63]

Answer:

B. +5.75 m/s

Explanation:

When there are two bodies, a and b, whose velocities measured by a third observer (in this case, the ground) are V_a and V_b respectively, the relative velocity of B with respect to A is given by:

V_{ba}=V_b-V_a

Thus, the velocity of the girl relative to the lawnmower is:

V_{ba}=6.5\frac{m}{s}-0.75\frac{m}{s}\\V_{ba}=5.75\frac{m}{s}

8 0
3 years ago
Read 2 more answers
A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal
Gemiola [76]

Answer:

Time taken, T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

T\ cos\theta-mg=0

T=\dfrac{mg}{cos\theta}

Sum of forces in x direction,

T\ sin\theta=\dfrac{mv^2}{r}

mg\ tan\theta=\dfrac{mv^2}{r}.............(1)

Also, r=l\ sin\theta

Equation (1) becomes :

mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}

v=\sqrt{gl\ tan\theta.sin\theta}...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

T=\dfrac{2\pi r}{v}

Put the value of T from equation (2) to the above expression:

T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}

T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}

On solving above equation :

T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Hence, this is the required solution.

4 0
3 years ago
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