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Valentin [98]
3 years ago
15

A small branch is wedged under a 200 kg rock and rests on a smaller object. The small object that creates a pivot point is calle

d the fulcrum. The smaller object is 2.0 m from the large rock and the branch is 12.0 m long. If the mass of the branch is negligible, what force must be exerted on the free end to just barely lift the rock
Physics
2 answers:
KiRa [710]3 years ago
8 0

Answer:

F = 326.7 N

Explanation:

given data

mass m = 200 kg

distance d = 2 m

length L = 12 m

solution

we know force exerted by the weight of the rock that is

W = m × g   ..............1

W = 200 × 9.8

W = 1960 N

and

equilibrium the sum  of the moment about that is

∑Mf = F(cos∅) L - W (cos∅) d  = 0

here ∅ is very small so cos∅ L = L and cos∅ d = D

so F × L - W × d = 0       .................2

put here value

F × 12 - 1960 × 2 = 0

solve it we get

F = 326.7 N

m_a_m_a [10]3 years ago
8 0

Answer:

F = 326.7 N

Explanation:

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So I don’t really get it can you also explain what’s the difference between radiation,conduction, and convection?
Slav-nsk [51]

Answer:

I think its radiation

Explanation:

Conduction is the transfer of heat through solids (A)

Convection is the transfer of heat through liquids or gasses (B)

Radiation is the transfer of heat through em waves (C)

5 0
3 years ago
A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis
Natali5045456 [20]

Answer:

A. 1.4 m/s to the left

Explanation:

To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:

M_{before} = M_{after}

where:

M = momentum [kg*m/s]

M = m*v

where:

m = mass [kg]

v = velocity [m/s]

(m_{1} *v_{1} )-(m_{2} *v_{2})=(m_{1} *v_{3} )+(m_{2} *v_{4})

where:

m1 = mass of the basketball = 0.5 [kg]

v1 = velocity of the basketball before the collision = 5 [m/s]

m2 = mass of the tennis ball = 0.05 [kg]

v2 = velocity of the tennis ball before the collision = - 30 [m/s]

v3 =  velocity of the basketball after the collision [m/s]

v4 = velocity of the tennis ball after the collision = 34 [m/s]

Now replacing and solving:

(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)

1 - (0.05*34) = 0.5*v3

- 0.7 = 0.5*v

v = - 1.4 [m/s]

The negative sign means that the movement is towards left

3 0
2 years ago
Belly-flop Bernie dives from atop a tall flagpole into a swimming pool below. His potential energy at the top is 7000 J (relativ
elena55 [62]

Answer:

KE₂ = 6000 J

Explanation:

Given that

Potential energy at top U₁= 7000 J

Potential energy at bottom U₂= 1000 J

The kinetic energy at top ,KE₁= 0 J

Lets take kinetic energy at bottom level =  KE₂

Now from energy conservation

U₁+ KE₁= U₂+ KE₂

Now by putting the values

U₁+ KE₁= U₂+ KE₂

7000+ 0 = 1000+ KE₂

KE₂ = 7000 - 1000 J

KE₂ = 6000 J

Therefore the kinetic energy at bottom is 6000 J.

5 0
3 years ago
Assuming a 8 kilogram bowling ball moving at 2 m/s bounces off a spring at the same speed that had before bouncing what is the a
Naya [18.7K]

a) 32 kg m/s

Assuming the spring is initially at rest, the total momentum of the system before the collision is given only by the momentum of the bowling ball:

p_i = m u = (8 kg)(2 m/s)=16 kg m/s

The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

p_f = p_{fB}+p_{fS}

where p_{fS} is the momentum of the spring. For the conservation of momentum,

p_i = p_f\\p_i = p_{fB}+p_{fS}\\p_{fS}=p_i -p_{fB}=16 kg m/s -(-16 kg m/s)=32 kg m/s


b) -32 kg m/s

The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:

\Delta p=p_{fb}-p_i=-16 kg m/s - 16 kg m/s=-32 kg m/s


c) 64 N

The change in momentum is equal to the product between the average force and the time of the interaction:

\Delta p=F \Delta t

Since we know \Delta t=0.5 s, we can find the magnitude of the force:

F=\frac{\Delta p}{\Delta t}=\frac{-32 kg m/s}{0.5 s}=-64 N

The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.


d) The force calculated in the previous step (64 N) is larger than the force of 32 N.

5 0
3 years ago
Answer all of these questions and you will get the brainlist
Rina8888 [55]
7) PE= Fwh = (72 m) (966 N) = 69552 Joules 7
8) zero PE=mass*g*height
I hope it’s correct
6 0
2 years ago
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