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3 years ago

All elements are composed of extremely small particles called

Physics

1 answer:

Tasya [4]3 years ago

8 0

The answer is Atoms they are in every single element

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A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat

Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

$\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}$

$\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}$

$\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}$

Using Newton's second law, it is established that F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object $\left(m_{1}+m_{2}\right)$

Now consider both $\mathrm{m}_{1} \text { and } \mathrm{m}_{2}$ as a system, so net force acting on the system is $\text { Force }=\left(m_{1}+m_{2}\right) a$

Substitute the given values in the above formula,

$\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

$\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0

4 years ago

Find the speed for Barkely, the dog, if he runs 5.5 miles in 3 hours

amm1812

Answer: B

Explanation:

5.5/3

6 0

3 years ago

Suppose a 96.10 kg hiker has ascended to a height of 1841 m above sea level in the process of climbing Mt. Washington. By what p

lbvjy [14]

Answer:

0.029%

Explanation:

Weight on the surface is given by,

$F_W = mg = (96.1Kg)(9.8m/s^2)=941.78N$

The acceleration due to gravity at given height is,

$g_h=\frac{gR^2}{R+h^2}$

The weight in another height is equal to

$F_h=mg_h$

$F_h= m*\frac{gR^2}{R+h^2}$

$F_h = \frac{F_WR^2}{(R+h)^2}$

$F_h=\frac{(941.78)(6.371*10^6)}{(6.371*10^6+1841)}$

$F_h = 941.50N$

The change of the Weight is,

$\frac{F-F_h}{F}*100=\frac{941.78-941.50}{941.78}*100=0.029\%$

6 0

3 years ago

Find the length of a microscope having magnifying power 100, if the focal lengths of the objective and eye piece are 2 cm and 5

sp2606 [1]

Answer:

L = 40 cm

Explanation:

A microscope is an optical instrument built by two lenses in such a way that the image of the first is formed within the distance of the other (eyepiece), so that the latter creates an enlarged virtual image of the object, for which the magnification of the microscope is the same to the multiplication of the magnification of each lens

M = - L / f₀ (25 cm / $f_{e}$ )

where fo and fe are the focal lengths of the objective and eyepiece, 25 cm is the near vision distance and L is the length of the microscope

L = - M f_{o} f_{e} / 25

let's calculate

L = - (-100) 2 5/25

L = 40 cm

7 0

3 years ago

If I measure the current to be 0.5 A and the voltage is 4.5 V. What is my Power?

Korvikt [17]

As we know that P=IV so P = (0.5A)(4.5V)
P= 2.25 watt

6 0

3 years ago