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otez555 [7]
2 years ago
9

A person is hauling their taco stand and it takes 3,500 Joules of work to stop the taco stand. What force was exerted on the tac

o stand if it took 1.5 meters to stop the motion?

Physics
1 answer:
Lapatulllka [165]2 years ago
3 0

Given the work done, the force exerted on the taco stand through the given distance is 2.3 × 10³ Newtons.

<h3>What is Work done?</h3>

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

W = f × d

Where f is force applied and d is distance travelled.

Given that;

  • Work done W = 3500J = 3500kgm²/s²
  • Distance covered d = 1.5m
  • Force applied F = ?

W = f × d

3500kgm²/s² = f × 1.5m

f = 3500kgm²/s² ÷ 1.5m

f = 2.3 × 10³ kgm/s²

f = 2.3 × 10³ N

Given the work done, the force exerted on the taco stand through the given distance is 2.3 × 10³ Newtons.

Learn more about work done here: brainly.com/question/26115962

#SPJ1

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An object has a coefficient of static friction of 0.3 and a normal force of 30 N. Find the force of static friction.
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Answer:

9N

Explanation:

static friction=normal force x coefficient of static friction

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3 years ago
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Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N==\frac{m(V)^{2}}{R}=m*(w)^2*R

Friction force\\Ff = k*N\\Ff= k*m*w^2*R

Gravitational force \\Fg = m*g

These must be balanced (the net force on the people will be 0) so set them equal to each other.

Fg = Ff

m*g = k*m*w^2*R

g=k*w^{2}*R

w^2 =\frac{g}{k*R}

w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}

There are 2*pi radians in 1 revolution so:

RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88

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3 years ago
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Explanation:

Mass of bumper cars, m_1=m_2=110\ kg

Initial speed of car A, u_1=8\ m/s

Initial speed of car Z, u_2=-10\ m/s

Final speed of car A after the collision, v_1=-10\ m/s

We need to find the velocity of car Z after the collision. Let it is equal to v_2. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

110\times 8+110\times (-10)=110\times (-10)+110v_2

v_2=\dfrac{-1320}{110}\ m/s

v_2=-12\ m/s

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

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