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3 years ago

Increase the mass of the beanbag its gravitational potential energy will

Physics

1 answer:

kotegsom [21]3 years ago

5 0

The gravitational potential energy will increase

Explanation:

The gravitational potential energy (GPE) of an object is the energy possessed by the object due to its position in a gravitational field.

Near the Earth's surface, the GPE of an object is given by

$GPE=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth of the object above the ground

From the equation, we see that the GPE is directly proportional to the mass: therefore, if the mass increases, the GPE will increase as well.

So, for the beanbag in this problem, when its mass increases, the GPE will increase as well.

Learn more about gravitational potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

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A 57 kg skier starts from rest at a height of H = 27 m above the end of the ski-jump ramp. As the skier leaves the ramp, his vel

Hatshy [7]

Answer:

(a) $h=5.95m$

(b) h is the same

Explanation:

According to the law of conservation of energy:

$E_i=E_f\\U_i+K_i=U_f+K_f$

The skier starts from rest, so $K_i=0$ and we choose the zero point of potential energy in the end of the ramp, so $U_f=0$ . We calculate the final speed, that is, the speed when the skier leaves the ramp:

$mgH=\frac{mv^2}{2}\\v=\sqrt{2gH}\\v=\sqrt{2(9.8\frac{m}{s^2})(27m)}\\v=23\frac{m}{s}$

Finally, we calculate the maximum height h above the end of the ramp:

$v_f^2=v_i^2-2gh\\$

The initial vertical speed is given by:

$v_i=vsin\theta$

and the final speed is zero, solving for h:

$h=\frac{v_i^2}{2g}\\h=\frac{((23\frac{m}{s})sin(28^\circ))^2}{2(9.8\frac{m}{s^2})}\\h=5.95m$

(b) We can observe that the height reached does not depend on the mass of the skier

3 0

4 years ago

PLEASE ANYONE CAN HELP ME !E.x/A block of metal has a volume of 0.09 m3

LuckyWell [14K]

Answer:

B = 1058.4 N

Explanation:

Given that,

The volume of a metal block, V = 0.09 m³

The density of fluid, d = 1200 kg/m³

We need to find the buoyant force when it's Completely immersed in brine. The formula for the buoyant force is given by :

$B=\rho gV$

g is acceleration due to gravity

$B=1200\times 9.8\times 0.09\\\\B=1058.4\ N$

So, the required buoyant force is 1058.4 N.

3 0

3 years ago

A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains

makkiz [27]

Answer:

3.834 m/s

Explanation:

h = 0.750 m

vx = 5 m/s

Let the initial vertical velocity is vy.

Final vertical velocity is zero

use third equation of motion along y axis

v^2 = vy^2 - 2 x g x h

0 = vy^2 - 2 x 9.8 x 0.75

vy^2 = 14.7

vy = 3.834 m/s

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4 years ago

A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them

Tatiana [17]

Answer:

A) q₂ = 75.98 cm, B) q₂' = 115.38 cm, C)

Explanation:

A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.

Lens 1. More to the left

let's use the constructor equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively,

We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.

$\frac{1}{q_1} = \frac{1}{f} - \frac{1}{p}$

$\frac{1}{q_1} = \frac{1}{14.8} - \frac{1}{50}$

1 / q₁ = 0.04756

q₁ = 21.0227 cm

this image is the object for the second lens that has f₂ = 14.8 cm

the distance must be measured from the second lens

p₂ = 39.4 -q₁

p₂ = 39.4 -21.0227

p₂ = 18.38 cm

let's use the constructor equation

1 / q₂ = 1 / f - 1 / p2

$\frac{1}{q_2} = \frac{1}{14.8} - \frac{1}{18.38}$

$\frac{1}{q_2}$ = 0.01316

q₂ = 75.98 cm

measured from the second lens

B) the position of the final image with respect to the first lens is

q₂’= q₂ + 39.4

q₂'= 75.98 +39.4

q₂' = 115.38 cm

C) the magnification of a lens is

m = - q / p

in this case the image measured from lens 2 is q2 = 75.98 cm

the distance to the object from the first lens is p1 = 50cm

m = - 75.98 / 50

m = -1.5 X

the negative sign indicates that the image is inverted

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The study of cosmos is of the large scale structures & evolution of the universe, it encompasses the origin, evolution, composition, and the eventual face of the universe.

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3 years ago

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