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Free_Kalibri [48]

3 years ago

5

Select the correct answer

2 answers:

Ainat [17]3 years ago

7 0

Answer:

the fulcrum is the part that the two levers meet.

Explanation:

gavmur [86]3 years ago

7 0

The answer is

D the fulcrum

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Which of these is an example of a mechanical wave? A. the light of the Sun B. a car stuck in traffic C. the shaking from an eart

sesenic [268]

A Explanation: I took this test before

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2 years ago

a car is traveling at velocity of 15m/s.it accelerates to a velocity 35m/s in 10s calculate the acceleration?

vlada-n [284]

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2m/s is acceleration I hope it helps you

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2 years ago

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A track is traveling a. a speed of 25.0 m/s along a level road. A crate is resting on the bed of the truck, and the coefficient

Mice21 [21]

To solve this problem it is necessary to apply the concepts related to

conservation of energy, for this case manifested through work and kinetic energy.

$W = \Delta KE$

$W = F*d$

Where,

F= Force (Frictional at this case $F_r = \mu N$ )

d= Distance

$\Delta KE = \frac{1}{2} mv^2$

Where,

m = mass

v = velocity

Equation both terms,

$F*d = \frac{1}{2}mv^2$

$\mu mg *d = \frac{1}{2}mv^2$

$\mu g * d = \frac{1}{2}v^2$

$d = \frac{1}{2} \frac{v^2}{\mu g}$

Replacing with our values we have that

$d = \frac{1}{2} \frac{25^2}{0.65*9.8}$

$d = 49.05m$

Therefore the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck is 49.05m

6 0

3 years ago

For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italia

Oliga [24]

Answer with Explanation:

We are given that

$m_1=0.3 kg$

$m_2=0.4 kg$

Spring constant, $k=200 N/m$

$h=0.250 m$

a.Speed of ham before collision

$u=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.25}=2.2m/s$

Collision is inelastic

According to law of conservation of momentum

$m_1u_1=(m_1+m_2)v$

$0.3\times 2.2=(0.3+0.4)v$

$v=\frac{0.3\times 2.2}{0.7}=0.94 m/s$

Kinetic energy of masses=Spring potential energy at maximum distance

$\frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}kA^2$

$A^2=\frac{(m_1+m_2)v^2}{k}$

$A=\sqrt{\frac{(m_1+m_2)v^2}{k}}$

Substitute the values

$A=\sqrt{\frac{(0.3+0.4)\times (0.94)^2}{200}}=0.056 m=0.056\times 100=5.6 cm$

1 m=100 cm

b.Time period, $T=2\pi\sqrt{\frac{(m_1+m_2)}{k}}$

$T=2\pi\sqrt{\frac{(0.3+0.4)}{200}}=0.37 s$

7 0

3 years ago

A book on a shelf 2.0 meters above the floor has a mass of 1.5 kilograms. What is the gravitational potential energy of the book

natta225 [31]

Answer:

29.4 J

Explanation:

PE= MGH

PE = 1.5 x 2x 9.8

4 0

3 years ago