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3 years ago

A forklift raises a crate weighing 8.35 × 102 newtons to a height of 6.0 meters. What amount of work does the forklift do?

Physics

2 answers:

Vinil7 [7]3 years ago

6 0

Answer:

5010J

Explanation:

Work is force times distance, or

W=F⋅d.

Substitute in values from the question to get

W=8.35⋅102N⋅6m=50.1⋅102Nm=5010J

Tatiana [17]3 years ago

4 0

Answer:

5.0*10^3 this is the answer. I swear it is

Explanation:

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Calculate the force of gravity between a comet with a mass of 500kg and a small asteroid with a mass of 20kg that is separated b

givi [52]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The equivalent gravitational force is ~

$F \approx1.48\times 10 {}^{ - 7} \: \: N$

$\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}$

We know that ~

$\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}$

where,

F = gravitational force

$m_1$ = mass of 1st object = 500 kg

$m_2$ = mass of 2nd object = 20kg

G = gravitational constant = $6.674 × {10}^ {-11}$

r = distance between the objects = 2.12 m

Let's calculate the force ~

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 500 \times 20}{(2.12) {}^{2} }$

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 10 {}^{4} }{4.4944}$

$F = \dfrac{6.674}{4.4944} \times 10 {}^{ - 7}$

$F =1.484 \times 10 {}^{ - 7} \: \: newtons$

7 0

2 years ago

The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for

abruzzese [7]

Answer:

The answer is " $\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}$ "

Explanation:

$\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta - \omega_{0}^{r} \\\\\to \omega^{r} = 2 (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}$

5 0

3 years ago

Which of the following substance are not formed by chemical bonds

Sliva [168]

Which of the following substance are not formed by chemical bonds? A MIXTURE

3 0

3 years ago

A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to

Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0

3 years ago

A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm

iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

$m = m_{o}* m_{e}$

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

$m_{o} = \frac{v_{o}}{-u _{0}}$

and me as

$m_{e} = 1+\frac{D}{f_{e}}$

d is distant of distinct vision = 25.0 cm for normal eye

fe = focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}$

$\frac{1}{v_{o}} = 2.1 cm$

$m_{o} = \frac{2.1}{0.150} = 14$

$m_{e} = 1+\frac{25}{1}$

$m_{e} =26$

$m = m_{o}* m_{e}$

m = 14*26 = 364

4 0

3 years ago