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Dmitry_Shevchenko [17]
2 years ago
6

A 35 kg boy is riding a 65 kg go-cart. He pushes on the gas pedal, causing the cart to accelerate at 5 m/s2. Use the equation F

= m × a to find the force exerted by the go-cart’s engine.
Physics
2 answers:
trasher [3.6K]2 years ago
4 0

Answer:

\boxed {\boxed {\sf 500 \ Newtons }}

Explanation:

The equation for force is given:

F=m*a

First, we must find the total mass, which is the sum of the boy's mass and the go-cart's mass.

  • total mass= boy's mass + go cart's mass

The boy's mass is 35 kilograms and the go cart's is 65 kilograms.

  • total mass= 35 kg+ 65 kg=100 kg

Now we know the total mass and the acceleration.

m= 100 \ kg \\a= 5 \ m/s^2

Substitute the values into the formula.

F=100 \ kg * 5 \ m/s^2

Multiply.

F= 500 \ kg*m/s^2

  • 1 kilograms meter per square second is equal to 1 Newton.
  • Our answer of 500 kg*m/s² is equal to 500 Newtons.

F= 500 \ N

The force exerted by the go cart engine is <u>500 Newtons.</u>

amid [387]2 years ago
4 0

Answer:

500 N

Explanation:

Your Welcome. (:

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1. A change in an object's speed has a(n) _________ effect on its kinetic energy than a change in its mass.
vredina [299]
A change in an object's speed has a(n) _________ effect on its kinetic energy than a change in its mass = <span>A greater effect.</span>
4 0
3 years ago
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A 6 m long, uniform ladder leans against a frictionless wall and makes an angle of 74.3 ◦ with the floor. The ladder has a mass
olganol [36]

Answer: µ=0.205

Explanation:

The horizontal forces acting on the ladder are the friction(f) at the floor and the normal force (Fw) at the wall. For horizontal equilibrium,

f=Fw

The sum of the moments about the base of the ladder Is 0

ΣM = 0 = Fw*L*sin74.3º - (25.8kg*(L/2) + 67.08kg*0.82L)*cos74.3º*9.8m/s²

Note that it doesn't matter WHAT the length of the ladder is -- it cancels.

Solve this for Fw.

0= 0.9637FwL - (67.91L)2.652

Fw=180.1/0.9637

Fw=186.87N

f=186.81N

Since Fw=f

We know Fw, so we know f.

But f = µ*Fn

where Fn is the normal force at the floor --

Fn = (25.8 + 67.08)kg * 9.8m/s² =

910.22N

so

µ = f / Fn

186.81/910.22

µ= 0.205

4 0
3 years ago
A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
  • m= 2,145 kg
  • v2= 12 \frac{m}{s}
  • v1= 25 \frac{m}{s}

Replacing:

W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0
3 years ago
Is a process that modifies light waves so they vibrate in a single plane
madam [21]

The process you're fishing for is "polarization", but that's a

misleading description.

Polarization doesn't do anything to change the light waves. 

It simply filters out (absorbs, as with a polarizing filter) the

light waves that aren't vibrating in the desired plane, and

allows only those that are to pass.

The intensity of a light beam is always reduced after

polarizing it, because much (most) of the original light

has been removed.

A laser light source may be thought of as an exception,

since everything coming out of the laser is polarized.

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Katarina [22]
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Heat required = mCΔT
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Heat required = 1.25 cal
3 0
3 years ago
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