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zhannawk [14.2K]
3 years ago
6

A force of 30n is applied to a screwdriver to pry the lid off of a can of paint. The screwdriver applies 75n of force to the lid

. What is the mechanical advantage of the screwdriver
Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0
The mechanical advantage of the screwdriver that is being described above is equal to 75N. This means that for every 30N that is applied on the screwdriver, this simple machine would in turn apply 75N of force to the lid of the can. 
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How much Tim in minutes will it take a car driving at 90km/hr to travel 27 kilometers
astraxan [27]

Answer:

18min

Explanation:

v=d/t

t=d/v= 27/90 =0.3hrs =18min

6 0
3 years ago
Uma massa de 500 Kg desloca-se com velocidade 58 km por hora. Calcule o módulo de sua quantidade por movimento
Simora [160]

The momentum of the object is 8050 kg m/s

Explanation:

The momentum of an object is defined as

p=mv

where

p is the momentum

m is the mass

v is the velocity of the object

For the object in this problem, we have

m = 500 kg is its mass

v = 58 km/h is its velocity

Converting the velocity into m/s,

v=58 \frac{km}{h}\cdot \frac{1000 m/km}{3600 s/h}=16.1 m/s

Therefore now we can find the momentum of the object:

p=(500)(16.1)=8050 kg m/s

Learn more about momentum:

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7 0
3 years ago
As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

Given that,

Height = 1.94  m

Bounced height = 1.48 m

Time interval t=14.86\times10^{-3}\ s

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

9.8\times1.94=\dfrac{1}{2}\times v^2

v=\sqrt{2\times9.8\times1.94}

v=6.166\ m/s

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

v'=\sqrt{2\times9.8\times1.48}

v'=5.38\ m/s

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

Put the value into the formula

a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

a=776.98\ m/s^2

a=0.77\times10^{3}\ m/s^2

Hence,The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

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