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3 years ago

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19) Which two structures are not found in animal cells? A) vacuoles and ribosomes B) chloroplasts and ribosomes C) cell walls an

d chloroplasts D) endoplasmic reticulum and cell walls

2 answers:

LUCKY_DIMON [66]3 years ago

7 0

The answer to your question is: C) Cell walls and chloroplasts

(I hope this helps!)

Molodets [167]3 years ago

5 0

Cell walls and Chloroplasts

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A copper penny has a mass of 3.0 g. A total of 4.0 × 1012 electrons are transferred from one neutral penny to another. If the el

sp2606 [1]

Answer:

F = K Q1 Q2 / R^2 = m g

Q1 = - Q2 = (4E12 * 1.60E-19) = 6.4E-7 coulombs

Q^2 = 4.10E-13

R^2 = 9.0E9 * 4.1E-13 / (.003 * 9.80) = 36.9E-4 / 2.94E-2

R^2 = .126

R = .35 m

5 0

2 years ago

During a hard sneeze, your eyes might shut for 0.32 s. If you are driving a car at 76 km/h during such a sneeze, how far does th

garik1379 [7]

Answer:

Distance, d = 6.75 meters

Explanation:

Given that,

Time taken by the eyes to shut during a hard sneeze, t = 0.32 s

Speed of the car, v = 76 km/h = 21.11 m/s

We need to find the distance covered by the car during this time. The product of speed and the time taken by the car is called the distance covered by the it. Mathematically,

$d=v\times t$

$d=21.11\times 0.32$

d = 6.75 meters

So, the car will cover 6.75 meters during that time. Hence, this is the required solution.

8 0

4 years ago

A 45.0-kg person steps on a scale in an elevator. The scale reads 460 n. What is the magnitude of the acceleration of the elevat

WINSTONCH [101]

Answer:

The magnitude of the acceleration of the elevator is 0.422 m/s²

Explanation:

Lets explain how to solve the problem

Due to Newton's Law ∑ Forces in direction of motion is equal to mass

multiplied by the acceleration

We have here two forces 460 N in direction of motion and the weight

of the person in opposite direction of motion

The weight of the person is his mass multiplied by the acceleration of

gravity

→ W = mg , where m is the mass and g is the acceleration of gravity

→ m = 45 kg and g = 9.8 m/s²

Substitute these values in the rule above

→ W = 45 × 9.8 = 441 N

The scale reads 460 N

→ F = 460 N , W = 441 N , m = 45 kg

→ F - W = ma

→ 460 - 441 = 45 a

→ 19 = 45 a

Divide both sides by 45

→ a = 0.422 m/s²

<em>The magnitude of the acceleration of the elevator is 0.422 m/s²</em>

4 0

4 years ago

A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes angle of 28.1 with the tensil

NISA [10]

Answer:

we have to find out the critical resolved shear stress. As it it given in the question

Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.

a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.

cos(62.4°) = 0.46

cos(72.0°) = 0.31

cos(81.1°) = 0.15

Thus, the slip direction is at the angle of 62.4° along the tensile axis.

b) now the critical resolved shear stress can be find out by the following equation.

τ $_{crss}$ = σ $_{Y}$ ( cosФ cosλ) $_{max}$

now by putting values,

= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23

3 0

4 years ago

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t

marishachu [46]

Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete ( $\Delta t$ ), in seconds:

$\Delta t = t_{A}-t_{C}$ (1)

Where:

$t_{C}$ - Time spent by the sound in concrete, in seconds.

$t_{A}$ - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

$\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}$

$\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}} \right)$

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$ (2)

Where:

$v_{C}$ - Speed of the sound in concrete, in meters per second.

$v_{A}$ - Speed of the sound in the air, in meters per second.

$x$ - Distance traveled by the sound, in meters.

If we know that $\Delta t = 6.4\,s$ , $v_{C} = 3000\,\frac{m}{s}$ and $v_{A} = 343\,\frac{m}{s}$ , then the distance travelled by the sound is:

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$

$x = 2478.585\,m$

The impact occured at a distance of 2478.585 meters from the person.

7 0

3 years ago