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1 year ago

5

Help me plz!

2 answers:

olga nikolaevna [1]1 year ago

5 0

As he lifts the sack to his chest from the floor

Inessa05 [86]1 year ago

4 0

Answer: Option (d) is the correct answer.

Explanation:

Work is defined as the force applied in order to move an object from a certain distance.

Mathematically, W = F × d

where, w = work

F = force

d = distance

Thus, when Lester lifts the sack to his chest from the floor then he is doing work as he is applying force on the sack in order to move it over his chest.

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If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be

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If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C.

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1 year ago

Read 2 more answers

Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree tha

emmasim [6.3K]

Answer:

The distance between knothole and the paint ball is 0.483 m.

Explanation:

Given that,

Height = 4.0 m

Distance = 15 m

Speed = 50 m/s

The angle at which the forester aims his gun are,

$\tan\theta=\dfrac{4}{15}$

$\tan\theta=0.266$

$\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}$

$\cos\theta=0.966$

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

$x=(u\cos\theta)t$

$t=\dfrac{x}{u\cos\theta}$

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

$y=u\sin\theta(t)-\dfrac{1}{2}gt^2$

$y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2$

$y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}$

Put the value into the formula

$y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})$

$y=3.517\ m$

We need to calculate the distance between knothole and the paint ball

$d=h-y$

$d=4-3.517$

$d=0.483\ m$

Hence, The distance between knothole and the paint ball is 0.483 m.

8 0

2 years ago

A metal bar moves through a magnetic field. the induced charges on the bar are

Dmitry [639]

I would say its a positive cgarge

7 0

2 years ago

A glider of mass m1 on a frictionless horizontal track is connected to an object of mass m2 by a massless string. The glider acc

inysia [295]

Answer:

m2g -> T2 -> T1

Explanation:

m2g -> T2 -> T1

5 0

1 year ago

If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol

xenn [34]

Answer:

Average density of Sun is 1.3927 $\frac{g}{cm}$ .

Given:

Radius of Sun = 7.001 × $10^{5}$ km = 7.001 × $10^{10}$ cm

Mass of Sun = 2 × $10^{30}$ kg = 2 × $10^{33}$ g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Solution:

Density of Sun is given by,

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Volume of Sun = $\frac{4}{3} \pi r^{3}$

Volume of Sun = $\frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}$

Volume of Sun = 1.436 × $10^{33}$ $cm^{3}$

Density of Sun = $\frac{ 2\times 10^{33} }{1.436 \times 10^{33} }$

Density of Sun = 1.3927 $\frac{g}{cm}$

Thus, Average density of Sun is 1.3927 $\frac{g}{cm}$ .

4 0

1 year ago