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Hoochie [10]
2 years ago
13

While a roofer is working on a roof that slants at 38.0 ∘ above the horizontal, he accidentally nudges his 95.0 n toolbox, causi

ng it to start sliding downward, starting from rest. part a if it starts 4.60 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 20.0 n ?

Physics
1 answer:
Stolb23 [73]2 years ago
7 0
1. Find all the forces on the toolbox
2. Use the total force to find the total acceleration
3. Use the acceleration to find the final velocity by combining these equations:
x =  \frac{1}{2} at^2 =\ \textgreater \  t^2 =  \frac{2x}{a}  \\ v = at =\ \textgreater \  t^2 =  \frac{v^2}{a^2}  \\ \\  v =   \sqrt{2xa}

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A proton starts from rest and gains 8.35 x 10^-14 joule of kinetic energy as it accelerates between points A and B in an electri
antoniya [11.8K]

Answer:

5.22 x 10^5 V

Explanation:

guessed on castle learning and got it right

6 0
2 years ago
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Of waterfalls with a height of more than 50 m , Niagara Falls in Canada has the highest flow rate of any waterfall in the world.
Vinil7 [7]

Answer:

Power output: W=1426.9MW

Explanation:

The power output of the falls is given mainly by its change in potential energy:

Q=-P_{tot}=-(P_{2}-P_{1})

The potential energy for any point can be calculated as:

P=m*g*h

If we consider the base of the falls to be the reference height, at point 2 h=0, so P2=0, and height at point 1 equals 52m:

Q=P_{1}=m*g*h

If we replace m with the mass rate M we obtain the rate of change in potential energy over time, so the power generated:

W=M*g*h=2.8*10^{3}m^{3}/s*1*10^{3}kg/m^{3}*9.8m/s^{2}*52m =1426.9MW

5 0
2 years ago
8. Rubbing a plastic bag and a balloon with a cloth gives both objects a net negative charge. The balloon's
Dafna1 [17]

Answer:

0.214 m

Explanation:

In order for the bag to levitate and not fall down, the electrostatic force between the bag and the balloon must balance the weight of the bag.

Therefore, we can write:

k\frac{q_1 q_2}{r^2}=mg

where

k is the Coulomb constant

q_1=-1\cdot 10^{-10}C is the charge on the balloon

q_2=-1\cdot 10^{-5} C is the charge on the bag

r is the separation betwen the bag and the balloon

m=0.02 g=2\cdot 10^{-5} kg is the mass of the bag

g=9.8 m/s^2 is the acceleration due to gravity

Solving for r, we find the distance at which the bag must be held:

r=\sqrt{\frac{kq_1 q_2}{mg}}=\sqrt{\frac{(9\cdot 10^9)(-1\cdot 10^{-10})(-1\cdot 10^{-5})}{(2\cdot 10^{-5})(9.8)}}=0.214 m

5 0
2 years ago
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

3 0
2 years ago
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B. velocity at position x, velocity at position x=0, position x, and the original position

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v_{x}^{2} = v_{ox}^{2} +2 a x (x - x₀)

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v_{ox} = velocity at position "x = 0 "

x = final position

x_{o} = initial position of the object at the start of the motion

6 0
2 years ago
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