Calculate the distance the gecko covers when sprinting from rest to 2 m/s in a time of 1.5 s.

In an arcade game a 0.099 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and releas

Sonja [21]

Answer:

The speed of disk is 1.98 $\frac{m}{s}$

Explanation:

Given:

Mass of $m = 0.099$ kg

Spring constant $k = 244 \frac{N}{m}$

Compression of spring $x = 4 \times 10^{-2}$ m

From energy conservation theorem,

Spring potential energy converted into kinetic energy,

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$

$v = \sqrt{\frac{k x^{2} }{m} }$

$v = \sqrt{\frac{244 \times 16 \times 10^{-4} }{0.099} }$

$v = 1.98 \frac{m}{s}$

Therefore, the speed of disk is 1.98 $\frac{m}{s}$

8 0

3 years ago

A certain digital camera having a lens with focal length 7.50 cmcm focuses on an object 1.70 mm tall that is 4.70 mm from the le

Crank

Answer:

a. 7.62cm

b. Real and inverted

c. 2.76 cm

d. 3450

Explanation:

We proceed as follows;

a. the lens equation that relates the object distance to the image distance with the focal length is given as follows;

1/f = 1/p + 1/q

making q the subject of the formula;

q = pf/p-f

From the question;

p = 4.70m

f = 7.5cm = 0.075m

Substituting these, we have ;

q = (4.7)(0.075)/(4.7-0.075) = 0.3525/4.625 = 0.0762 = 7.62 cm

b. The image is real and inverted since the image distance is positive

c. We want to calculate how tall the image is

Mathematically;

h1 = (q/p)h0

h1 = (7.62/4.70)* 1.7

h1 = 2.76 cm

d. We want to calculate the number of pixels that fit into this image

Mathematically:

n = h1/8 micro meter

n = 2.76cm/8 micro meter = 2.76 * 10^-2/8 * 10^-6 = 3450

3 0

3 years ago

Una persona de 80 kg de masa corre a una velocidad cuya magnitud es de 9m/s. Cual es la magnitud de la cantidad de movimiento? Q

natka813 [3]

Responder:

13,01 m / s

Explicación:

Paso uno:

datos dados

masa de la persona 1 m = 80 kg

velocidad de la persona 1 v = 9 m / s

masa de la persona 2 M = 55kg

velocidad de la persona 2 v =?

Segundo paso:

la expresión del impulso se da como

P = mv

para la primera persona, el impulso es

P = 80 * 9

P = 720N

Paso tres:

queremos que la segunda persona tenga el mismo impulso que la primera, por lo que la velocidad debe ser

720 = 55v

v = 720/55

v = 13,09

v = 13,01 m / s

Por lo tanto, la magnitud de la velocidad debe ser 13.01 m / s.

4 0

3 years ago

When the spacecraft is at the halfway point, how does the strength of the gravitional force on the spaceprobe by Earth compre wi

mixer [17]

Solution :

When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.

The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.

5 0

3 years ago

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0

3 years ago