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3 years ago

Explain why some objects fall faster than other objects,

1 answer:

8 0

objects that are more dense, or have more mass, fall at a faster rate than less dense objects, due to this air resistance. A feather and brick dropped together. Air resistance causes the feather to fall more slowly.

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What is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?

Art [367]

CORRECT ANSWER:

a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.

STEP-BY-STEP EXPLANATION:

The complete question from book is

According to Figure 9.6, what is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?

a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.

b- Signaling molecules that bind to cell-surface receptors lead to cellular responses restricted to the cytoplasm; signaling molecules that bind to intracellular receptors lead to cellular responses restricted to the nucleus.

c- Cell-surface receptors bind to specific signaling molecules; intracellular receptors bind any signaling molecule.

d- Cell-surface receptors typically bind to signaling molecules that are smaller than those bound by intracellular receptors.

e- None of the other answer options is correct.

3 0

3 years ago

Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth

zalisa [80]

Refer to the diagram shown below.

Given:
R = 6.37 x 10⁶ m, the radius of the earth
h = 3.58 x 10⁷ m, the height of the satellite above the earth's surface.
Therefore
R + h = 4.217 x 10⁷ m

In geosynchronous orbit, the period of rotation is 1 day.
Therefore the period is
T = (24 h)*(60 min/h)*(60 s/min) = 86400 s

The angular velocity is
ω = (2π rad)/(86400 s) = 7.2722 x 10⁻⁵ rad/s

Part (a)
The tangential speed is
v = (R+h)*ω
= (4.217 x 10⁷ m)*(7.2722 x 10⁻⁵ rad/s)
= 3066.7 m/s
= 3.067 km/s

Part (b)
The centripetal acceleration is
a = v²/(R+h)
= (3066.7 m/s)²/(4.217 x 10⁷ m)
= 0.223 m/s²

Answers:
(a) The speed is 3.067 km/s
(b) The acceleration is 0.223 m/s²

7 0

3 years ago

PLS HELP!!

stealth61 [152]

Answer:

Explanation:

The iris controls the amount of light that enters the eye by opening and closing the pupil. The iris uses muscles to change the size of the pupil. These muscles can control the amount of light entering the eye by making the pupil larger (dilated) or smaller (constricted).

8 0

2 years ago

Read 2 more answers

A pharmacist attempts to weigh 100 milligrams of codeine sulfate on a balance with a sensitivity requirement of 4 milligrams. Ca

ira [324]

Answer:

4 %

2 ) 3.42 %

Explanation:

Sensitivity requirement of 4 milligram means it is not sensitive below 4 milligram or can not measure below 4 milligram .

Given , 4 milligram is the maximum error possible .

Measured weight = 100 milligram

So percentage maximum potential error

= (4 / 100) x 100

4 %

2 )

As per measurement

weight of 6 milliliters of water

= 48.540 - 42.745 = 5.795 gram

6 milliliters of water should measure 6 grams

Deviation = 6 - 5.795 = - 0.205 gram.

Percentage of error =(.205 / 6 )x 100

= 3.42 %

3 0

3 years ago

A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the

olga2289 [7]

Answer:

a) F = 30 N, b) I = 12 N s , c) I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

p₀ = m v

The speed can be found by kinematics

v = v₀ - g t

v = 0 - 10 0.4

v = -4.0 m / s

Final after division

pf = m₁ v₁f + m₂ v₂f

p₀ = pf

M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

F = mg

F = 3 10 = 30 N

b) The expression for momentum is

I = Ft

Before the explosion the only force that acts is the weight

I = mg t

I = 3 10 0.4

I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

M v = m₁ v₁f + m₂ v₂f

v₂f = (M v - m₁ v₁f) / m₂

v₂f = (3 (-4) - 1 6) / 2

v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

I = F t = Dp

F t = pf –po)

F t= [m₁ v₁f + m₂ v₂f]

I = [1 6 + 2 (-9) -0]

I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

ΔI =If – I₀

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0

3 years ago