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Komok [63]
3 years ago
7

If a lawn mower is pushed with a distance of 30 meters and 12N-m of work is exerted, calculate the force.

Physics
1 answer:
Nimfa-mama [501]3 years ago
7 0

Answer:

Explanation:

W = FΔx so filling in:

12 = F(30) so

F = .4N

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A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
2 years ago
What is a passing mechanic cues to remember when you are the receiver of a pass in hockey?
VARVARA [1.3K]

Answer:

Positions in Hockey: 6 players for each team on the ice

1 Goalie – the player in the goal who tries to stop the puck from going in the net.

1 Center – plays in between the two wings and is usually the best passer on the team

2 Wings – offensive players who plays on both sides of the center. They are usually goal scorers

2 Defensemen – main job is to play defense and help defend the goal

Passing Cues

1. Stick blade faces target

2. Puck in center of blade

3. Transfer weight rear to front as you pass

4. Use wrist movement to drive the puck

5. Follow through at target

Receiving Cues:

1. athletic position

2. catch puck with middle of blade and control

3. slow the puck when it contacts the stick by giving with it

Explanation:

6 0
2 years ago
Is the wavelength comparable to the size of atoms?
Helen [10]

It totally depends on what kind of wave you're talking about.

-- a sound wave from a trumpet or clarinet playing a concert-A pitch is about 78 centimeters long ... about 2 and 1/2 feet.  This is bigger than atoms.

-- a radio wave from an AM station broadcasting on 550 KHz, at the bottom of your radio dial, is about 166 feet long ... maybe comparable to the height of a 10-to-15-story building.  This is bigger than atoms.

-- a radio wave heating the leftover meatloaf inside your "microwave" oven is about 4.8 inches long ... maybe comparable to the length of your middle finger.  this is bigger than atoms.

-- a deep rich cherry red light wave ... the longest one your eye can see ... is around 750 nanometers long.  About 34,000 of them all lined up will cover an inch.  These are pretty small, but still bigger than atoms.

-- the shortest wave that would be called an "X-ray" is 0.01 nanometer long.     You'd have to line up 2.5 billion of <u>those</u> babies to cover an inch.  Hold on to these for a second ... there's one more kind of wave to mention.

-- This brings us to "gamma rays" ... our name for the shortest of all electromagnetic waves.  To be a gamma ray, it has to be shorter than 0.01 nanometer.

Talking very very very very roughly, atoms range in size from about 0.025 nanometers to about 0.26 nanometers.

The short end of the X-rays, and on down through the gamma rays, are in this neighborhood.

5 0
3 years ago
If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m
AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}

We can cancel the common factor of m which leaves us with

gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}

Lets solve for v_f

gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}

Subtract gh_f from both sides of the equation.

gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}

Multiply both sides of the equation by 2.

2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}

Simplify the left side.

Apply the distributive property.

2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}

Cancel the common factor of 2.

2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}

Take the square root of both sides of the equation to eliminate the exponent on the right side.

{v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}

We are given g,v_{i},h_{i},h_{f}.

We can now solve for the final velocity.

{v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)

Anything multiplied by 0 is 0.

{v_{f}=\sqrt{2*9.81*2.41

{v_{f}=\sqrt{47.2842

v_f=6.88

7 0
1 year ago
If an athlete leaps vertically at 4.0m/s, what maximum height does he reach?
joja [24]
Hello
This is a problem of accelerated motion, where the acceleration involved is the gravitational acceleration: g=-9.81~m/s^2, and where the negative sign means it points downwards, against the direction of the motion.

Therefore, we can use the following formula to solve the problem:
v_f^2 = v_i^2 + 2gS
where v_i=4~m/s is the initial vertical velocity of the athlete, v_f=0 is the vertical velocity of the athlete at the maximum height (and v_f=0~m/s at maximum height of an accelerated motion) and S is the distance covered between the initial and final moment (i.e., it is the maximum height). Re-arranging the equation, we get
S= \frac{v_f^2-v_i^2}{2g}=0.82~m

3 0
3 years ago
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