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2 years ago

A third class lever cannot magnify force why

Physics

1 answer:

seropon [69]2 years ago

6 0

Answer:

The third class lever cannot magnify our force

Explanation:

because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.

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A 92-kg skier is sliding down a ski slope that makes an angle of 30 degrees above the horizontal direction. The coefficient of k

9966 [12]

Answer:

a = 4.05 m/s²

Explanation:

Known data

m= 92 kg : mass of the skier

θ =30° :angle θ of the ski slope with respect to the horizontal direction

μk= 0.10 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the skier

W: Weight of the skier : In vertical direction

N : Normal force : perpendicular to the ski slope

f : Friction force: parallel to the ski slope

Calculated of the W

W= m*g

W= 92kg* 9.8 m/s² = 901,6 N

x-y weight components

Wx= Wsin θ= 901,6 N *sin 30° = 450.8 N

Wy= Wcos θ = 901,6 N *cos 30° =780.8 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay ay = 0

N - Wy = 0

N = Wy

N = 780.8 N

Calculated of the f

f = μk* N= 0.10*780.8 N

f = 78.08 N

We apply the formula (1) to calculated acceleration of the skier:

∑Fx = m*ax , ax= a : acceleration of the block

Wx - f = m*a

450.8- 78.08 = ( 92)*a

372.72 = (92)*a

a = (372.72)/ (92)

a = 4.05 m/s²

6 0

2 years ago

When a pendulum is pulled back from its equilibrium position by 10∘, the restoring force is 1.0 N. When it is pulled back to 30∘

jarptica [38.1K]

Answer: B

Explanation: I said B because if you pull something back what is going to be more of a force pulling back or letting it go for a rubier band yes it will have more force if you let it go

5 0

2 years ago

Read 2 more answers

A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How lon

frosja888 [35]

Answer:

t=0.704s

Explanation:

A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How long will it take the car to reach the bottom of the ramp if there is no friction?

from newton equation of motion , we look for the y component of the speed and look for the x component of the speed. we can then find the resultant of the speed

$V^{2} =u^{2} +2as$

Vy^2=0+2*9.8*1.73sin40.5

Vy^2=22.021

Vy=4.69m/s

Vx^2=u^2+2*9.81*cos40.5

Vy^2=25.81

Vy=5.08m/s

V=(Vy^2+Vx^2)^0.5

V=47.71^0.5

V=6.9m/s

from newtons equation of motion we know that force applied is directly proportional to the rate of change in momentum on a body.

f=force applied

v=velocity final

u=initial velocity

m=mass of the toy, 0.046

f=ma

f=m(v-u)/t

v=u+at

6.9=0+9.8t

t=6.9/9.81

t=0.704s

4 0

2 years ago

Explain why the brakes of a car get much hotter than the brakes of a bicycle?

ValentinkaMS [17]

Answer: Car brakes produces more energy then the bicycle because the cars wheels produces a much bigger force that makes the car go and to stop that force the car uses greater amount of energy that transfers to heat but in a bicycle the wheels do not turn that fast so when you press the break there is less energy that transfer to heat.

Explanation:

8 0

2 years ago

The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha

saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0

2 years ago

A third class lever cannot magnify force why​

A third class lever cannot magnify force why