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Bogdan [553]
2 years ago
7

A screwdriver is used to pry the lid off a paint can. The resistance arm is 0.5 cm long and the effort arm is 20 cm long. What i

s the mechanical advantage of the screwdriver?
Physics
1 answer:
Lunna [17]2 years ago
4 0

Answer:

40

Explanation:

Mechanical advantage = effort arm / load arm

MA = 20 cm / 0.5 cm

MA = 40

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4

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The two major problems with most motor vehicles are that they burn fossil fuels and _____________.
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A. Create radioactive waste i believe

Explanation:

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You and your family take a trip to see your aunt who lives 100 miles away along a straight highway. The first 60 miles of the tr
Nitella [24]

Answer:

51.2 mi/h

Explanation:

Total distance, d = 100 miles

First 60 miles with speed 55 mi/h

Next 40 miles with speed 75 mi/h

Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h

Time taken for 40 miles, t2 = 40 / 75 = 0.533 h

Time spent to get stuck, t3 = 20 min = 0.33 h

Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h

The average speed is defined as the ratio of total distance traveled to the total time taken.

Average speed = =\frac{100}{1.953}=51.2 mi/h

Thus, the average speed of the journey is 51.2 mi/h.

4 0
2 years ago
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2000 kg, was stopped at a red light w
OLga [1]

Answer:

a). va=17.23 \frac{m}{s} or 38.54 mph

b). v=38.54 mph and limit is 35 mph

c). Completely inelastic

d). Eka=192.967 kJ

Ekt=76.071 kJ

Explanation:

m_{a}=1300kg\\m_{b}=2000kg\\x_{f}=7.25m\\u_{k}=0.65

The motion is an inelastic collision so

m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}

The force of the motion is contrarest by the force of friction so

F-F_{uk} =0\\F=F_{uk}\\F_{uk}=u_{k}*m*g\\F=m*a\\a=\frac{F}{m}\\ a=\frac{F_{uk}}{m}\\a=\frac{u_{k}*m*g}{m}\\a=u_{k}*g\\a=0.65*9.8\frac{m}{s^{2}} \\a=6.39\frac{m}{s^{2}}

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

x_{f}=x_{o}+v_{o}*t+2*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=2*a*t^{2}\\t^{2}=\frac{x_{f}}{2*a}\\t=\sqrt{\frac{7.25m}{6.37\frac{m}{s^{2} } } } \\t=1.06s

So the velocity final can be find using this time

v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=6.37\frac{m}{s^{2} } *1.06s\\v_{f}=6.79 \frac{m}{s}

a).

Replacing in the first equation the final velocity can find the initial velocity

m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}

v_{b}=0

v_{a}= \frac{(m_{a}+m_{b)*v_{f}}}{m_{a}}\\v_{a}= \frac{(1300+2000)*6.37}{1300}\\v_{a}=17.23 \frac{m}{s}

b).

35mph*\frac{1m}{0.000621371mi} *\frac{1h}{3600s}=15.646\frac{m}{s}

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s

so is exceeding the speed limit in about 1.58 m/s

or in miles per hour

3.5 mph

c).

The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic

d).

Ek=\frac{1}{2}*m*(v)^{2}\\  Eka=\frac{1}{2}*1300kg*(17.23\frac{m}{s})^{2}\\Eka=192.967 kJ\\Ekt=\frac{1}{2}*m*(v)^{2}\\Ekt=\frac{1}{2}*3300kg*(6.79\frac{m}{s})^{2}\\Ekt=76.071 kJ

8 0
2 years ago
I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a
attashe74 [19]
<span>The angular momentum of a particle in orbit is 

l = m v r 

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2" 

m_1 v_1 r_1 = m_2 v_2 r_2 

Assuming that the mass did not change, conservation of angular momentum demands that 

v_1 r_1 = v_2 r_2 

or 

v1 = v_2 (r_2/r_1) 

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
3 0
2 years ago
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