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3 years ago

The long handle on a rake is a(n) _____.

Physics

2 answers:

garik1379 [7]3 years ago

8 0

The correct answer is A. Lever

Zigmanuir [339]3 years ago

4 0

In my opinion, I believe it’s A(lever)

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A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that

lina2011 [118]

Answer:

0.006075Joules

Explanation:

The final kinetic energy of the system is expressed as;

KE = 1/2(m1+m2)v²

m1 and m2 are the masses of the two bodies

v is the final velocity of the bodies after collision

get the final velocity using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

0.12(0.45) + 0/12(0) = (0.12+0.12)v

0.054 = 0.24v

v = 0.054/0.24

v = 0.225m/s

Get the final kinetic energy;

KE = 1/2(m1+m2)v

KE = 1/2(0.12+0.12)(0.225)²

KE = 1/2(0.24)(0.050625)

KE = 0.12*0.050625

KE = 0.006075Joules

Hence the final kinetic energy of the system is 0.006075Joules

5 0

2 years ago

A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori

olasank [31]

Explanation:

The given data is as follows.

k = 130 N/m, $\Delta x$ = 17 cm = 0.17 m (as 1 m = 100 cm)

mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

Energy = $\frac{1}{2}kx^{2}$

Now, spring energy gets converted into kinetic energy when the box is launched.

So, $\frac{1}{2}kx^{2}$ = $\frac{1}{2}mv^{2}$

$\frac{1}{2} \times 130 \times (0.17)^{2}$ = $\frac{1}{2} \times 2.8 \times v^{2}$

$v^{2} = \frac{3.757}{2.8}$

= 1.34

v = 1.15 m/sec

Now,

Frictional force = $\mu \times mg$

= $0.15 \times 2.8 \times 9.8$

= 4.116 N

Also, Kinetic energy = work done by friction

$\frac{1}{2}mv^{2} = F_{f} \times d$

$\frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d$

1.8515 = $4.116 \times d$

d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0

3 years ago

A camera is equipped with a lens with a focal length of 34 cm. When an object 2.4 m (240 cm) away is being photographed, what is

puteri [66]

Answer:

The magnification is -6.05.

Explanation:

Given that,

Focal length = 34 cm

Distance of the image =2.4 m = 240 cm

We need to calculate the distance of the object

$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$

Where, u = distance of the object

v = distance of the image

f = focal length

Put the value into the formula

$\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}$

$\dfrac{1}{u}=\dfrac{103}{4080}$

$u =\dfrac{4080}{103}$

The magnification is

$m = \dfrac{-v}{u}$

$m=\dfrac{-240\times103}{4080}$

$m = -6.05$

Hence, The magnification is -6.05.

6 0

3 years ago

sunlight, wind, and running water are essentially "free." Yet renewable energy sources are a very small part of our energy consu

dimulka [17.4K]

The heat and energy of coal burning is more easier and more efficient to turn into electricity than any method using renewable resources.

7 0

3 years ago

Read 2 more answers

What is -8,675,309.0 in scientific notation

Sindrei [870]

= 3.456 × 1011
(scientific notation)

= 3.456e11
(scientific e notation)

= 345.6 × 109
(engineering notation)
(billion; prefix giga- (G))

= 345600000000
(real number)

4 0

3 years ago