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3 years ago

Alice kicks a 0. 25 kg soccer ball with 0. 5 N of force. What force does the ball exert on Alice’s foot as she kicks it? N.

1 answer:

3 0

The force exerted on Alice’s foot as she kicks the ball is 1.25 N.Force can be defined as the influence that can accelerate the object.

<h3>What is force?</h3>

Force can be defined as the influence that can accelerate the object. From Newton's Second Law of motion,

$F = ma$

Where,

$F$ - force

$m$ - mass = 0. 25 kg

$a$ - acceleration = 5 N

Put the values in the formula,

$F = 0.25 \times 5\\\\F = 1.25 \rm \ N$

Therefore, the force exerted on Alice’s foot as she kicks the ball is 1.25 N.

Learn more about Force:

brainly.com/question/2840101

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Explanation:

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3 years ago

In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re

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Answer:

$I=2.6363\ kg.m^2$

Explanation:

Given:

dimension of uniform plate, $(0.673\times 0.535)\ m^2$

mass of plate, $m=10.7\ kg$

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

$I_{cm}=\frac{1}{12} \times m(L^2+B^2)$

where:

$L=$ length of the plate

$B=$ breadth of the plate

$I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)$

$I_{cm}=0.6591\ kg.m^2$

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

$s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}$

$s=0.4299\ m$

Now using parallel axis theorem:

$I=I_{cm}+m.s^2$

$I=0.6591+10.7\times 0.4299^2$

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3 years ago

Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t

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Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

$y=\frac{\lambda (m+\frac{1}{2})D}{d}$

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light used

In this problem we have:

$\lambda=683 nm = 683\cdot 10^{-9} m$ is the wavelength of the light

$d=1.1 mm = 0.0011 m$ is the width of the slit

m = 13 is the order of the minimum

$y=8.57 cm = 0.0857 m$ is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

$D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m$

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3 years ago

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A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the

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We use the equation of motion for vertical component,

$s_{y} = u_{y} t+\frac{1}{2} gt^2.$

Here, $s_{y}$ is displacement of bullet, $u_{y}$ is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

$s_{y} =\frac{1}{2}gt^2$

Given, $s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m$

Substituting the values, we get time of flight

$2.3 \times 10^{-2} \ m = \frac{1}{2} \times 9.8 \ m/s^2 \times t^2 \\\\ t =\sqrt{46.94 \times 10^{-4} \ s } = 0.069 \ s$

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3 years ago