Which of the following types of models could be used to represent a cell?

At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m

eimsori [14]

Answer:

1) The time it takes the diver to move off the end of the diving board to the pool surface, $t_w$ , is approximately 1.392 seconds

2) The horizontal distance from the edge of the pool to where the diver enters the water, $d_w$ , is approximately 5.76 meters

Explanation:

1) The given parameters are;

The height of the diving board above the pool's surface, h = 9.5 m

The length by which the diving board over hangs the pool L = 2 m

The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s

Taking $t_w$ = The time it takes the diver to move off the end of the diving board to the pool surface

Therefore, we have from the equation of free fall;

h = 1/2 × g × $t_w$ ²

Where;

g = The acceleration due to gravity = 9.81 m/s²

Substituting the values, gives;

9.5 = 1/2 × 9.81 × $t_w$ ²

$t_w$ = √(9.5/(1/2 × 9.81)) ≈ 1.392 s

The time it takes the diver to move off the end of the diving board to the pool surface = $t_w$ ≈ 1.392 s

2) The horizontal distance, $d_w$ , in meters from the edge of the pool to where the diver enters the water is given as follows;

$d_w$ = L + v₀ × $t_w$ = 2 + 2.7× 1.392 ≈ 5.76 m

∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.

7 0

3 years ago

You weigh 716 newtons on Earth. You

Otrada [13]

Answer:

537 N

Explanation:

The force due to gravity of a planet is:

F = GMm / r²

where G is the universal gravitational constant

M is the mass of the planet

m is the mass of the object

and r is the distance between the object and the center of the planet

On Earth, you weigh 716 N, so:

716 N = GMm / r²

On planet X:

F = G (3M) m / (2r)²

F = 3/4 GMm / r²

F = 3/4 (716 N)

F = 537 N

8 0

4 years ago

In a simulation on earth, an astronaut in his space suit climbs up a vertical ladder. On the moon, the same astronaut makes the

mr Goodwill [35]

Answer:

Gravitational potential energy of the astronaut will change by a greater amount on the earth

Explanation:

Gravitational potential energy is expressed by the formula;

GPE = mgh

This means that the gravitational potential energy is directly proportional to the gravity(g)

Now, from constant values, gravity of moon is 1.62 m/s² while gravity of the earth is 9.81 m/s².

This means that if we plug in the values of g on the earth and g on the moon, the potential energy on the earth would be greater than that of the moon

Thus, gravitational potential energy of the astronaut will change by a greater amount on the earth

8 0

3 years ago

A solid conducting sphere has net positive charge and radius r = 0.800 m . At a point 1.20 m from the center of the sphere, the

Serga [27]

here we have to calculate the net positive charge present on he surface of the conducting sphere.

as the sphere is a conducting one the,the charge will be stored on its surface.

though the charge is present at the surface,still it will behave just like the total charge is concentrated at the centre of the sphere.

the electric field at outside of the sphere is E= $\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

as E= $\frac{-dv}{dr}$

so V= $\frac{1}{4\pi\epsilon} \frac{q}{r}$

here V=27 volt,radius of sphere is 0.800 m and the point at which the potential is considered is at 1.20 m from th centre of the sphere.hence the point is at the outside of the sphere .

hence we have 27= $\frac{1}{4\pi\epsilon} \frac{q}{1.20}$

q=27× $\frac{1}{9*10^{9} }* 1.20$ coulomb

=3.6× $10^{-9}$ coulomb [ans]

5 0

3 years ago

Consider a metal single crystal oriented such that the normal to the slip plane and the slip directions are at angles of 43.1° a

Bond [772]

Answer:

resolve shear stress = 22 MPa

Explanation:

Given data

slip plane α = 43.1°

slip directions β = 47.9°

shear stress = 20.7 MPa (3,000 psi)

applied stress =45 mPa (6,500 psi)

to find out

what stress will be necessary

solution

we know that

resolve shear stress = aplied stress × cosα × cosβ

resolve shear stress = 45 × cos(43.1) × cos(47.9)

resolve shear stress = 22 MPa

we can say that here single cristal will be yield

because resolve shear stress is bigger than critical shear stress

3 0

4 years ago