Calculate the force of gravity between two objects of masses 1300 kg and 7800 kg, which are 0.23 m apart.

The steel plate is 0.3 m thick and has a density of 7850 kg>m3 . determine the location of its center of mass. also find the

cestrela7 [59]

answer with full explanation is attached below

3 0

1 year ago

For sprinters running at 12 m/s around a curved track of radius 26 m, how much greater (as a percentage) is the average total fo

Snezhnost [94]

Answer:

114.86%

Explanation:

In both cases, there is a vertical force equal to the sprinter's weight:

Fy = mg

When running in a circle, there is an additional centripetal force:

Fx = mv²/r

The net force is found with Pythagorean theorem:

F² = Fx² + Fy²

F² = (mv²/r)² + (mg)²

F² = m² ((v²/r)² + g²)

F = m √((v²/r)² + g²)

Compared to just the vertical force:

F / Fy

m √((v²/r)² + g²) / mg

√((v²/r)² + g²) / g

Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:

√((12²/26)² + 9.8²) / 9.8

1.1486

The force is about 114.86% greater (round as needed).

5 0

1 year ago

--->Two aircraft P and Q are flying at the same speed. 300 m/s, The direction along which P is flying is at right angles to t

REY [17]

Answer:

The magnitude of the velocity of the aircraft P relative to aircraft Q is zero

Explanation:

The velocity of the two aircraft, P & Q, v = 300 m/s

The angle of the direction between them, Ф = 90°

The magnitude of the velocity of aircraft P relative to aircraft Q is given by the formula

Substituting the values in the above equation

v = 300 x cos 90°

= 300 x 0

= 0

Since the aircraft are at right angles, the velocity of one aircraft relative to the other is zero.

5 0

2 years ago

A frictionless inclined plane is 8.0 m long and rests on a wall that is 2.0 m high. How much force is needed to push a block of

adoni [48]

Given Information:

Inclined plane length = 8 m

Inclined plane height = 2 m

Weight of ice block = 300 N

Required Information:

Force required to push ice block = F = ?

Answer:

Force required to push ice block = 75 N

Explanation:

The force required to push this block of ice on a inclined plane is given by

F = Wsinθ

Where W is the weight of the ice block and θ is the angle as shown in the attached image.

Recall from trigonometry ratios,

sinθ = opposite/hypotenuse

Where opposite is height of the inclined plane and hypotenuse in the length of the inclined plane.

sinθ = 2/8

θ = sin⁻¹(2/8)

θ = 14.48°

F = 300*sin(14.48)

F = 75 N

Therefore, a force of 75 N is required to push this ice block on the given inclined plane.

7 0

2 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

2 years ago