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4 years ago

Which of the following is not true of mutations?

2 answers:

8 0

The answer is
A Mutations can't result in a change in appearance
because quit contrary they can for example what would be a white mouse with a mutation in fur color can have brown fur due to the mutation

igomit [66]4 years ago

6 0

Answer:

A. Mutations can't result in a change in appearance.

Explanation:

A mutation is a sudden change in DNA that occurs at random in genetic material and can be transmitted to offspring. A mutation can also be caused by exposure to ultraviolet or ionizing radiation, chemical mutagens, or viruses. When the mutation occurs, it is permissible and continues with the individual until the day of death, as it is irreversible as it has modified the coding of DNA. When a mutation occurs, it can cause differences in the appearance of the individual, such as Down's syndrome, which changes not only the appearance, but the organism altogether. Therefore, we can consider that the answer to your question is the letter A.

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A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is

Maru [420]

Answer:

A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is the surface charge density of the sheet"?

Explanation:

Surface charge density is a measure of how much electric charge is accumulated over a surface. It can be calculated as the charge per unit area.

We will convert all parameters in SI units.

Charge = Q = -5.02nC

Q = -5.02× $10^{-9}$ C

As it is clear from question that Sheet is a square (All sides will be of equal length)

Area = A = (21.8× $10^{-2}$ m) (21.8× $10^{-2}$ m) = 4.75× $10^{-4}$ m²

A = 4.75× $10^{-4}$ m²

Surface charge density = Q/A

Surface charge density = (-5.02× $10^{-9}$ C)/(4.75× $10^{-4}$ m²)

Surface charge density = -1.057× $10^{-5}$ C $m^{-2}$

3 0

3 years ago

Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s

alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
vₓ₀ = v * cos θ (1)
where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

$x_{f} = v_{ox} * t = v_{o} * cos \theta * t (2)$

Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

$cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)$

⇒θ = cos⁻¹ (0.526) = 58.3º (4)

At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

$\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)$

Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
Replacing (7) in (6), we get:

$\Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)$

When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
The horizontal component, since it keeps constant, is just v₀x:
v₀ₓ = 13.7 m/s
The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

$v_{fy} = v_{oy} - g*t (9)$

Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:

$v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s (10)$

Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

$v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)$

3 0

3 years ago

A block oscillating up and down on a spring comes to the top of its path, where it is momentarily at rest. At the instant it is

True [87]

Opposite to the direction of the velocity which led it to its current position.

Explanation:

The direction of momentum when a vertically oscillating block comes to the rest momentarily will be opposite to the direction of the velocity that it has just followed to reach reach its current position.

The direction of change in momentum at the bottom will be upwards and at the top will be downwards.

The change in momentum is mathematically defined as:

$\Delta P=m.v_f-m.v_i$

where:

$m=$ mass of the block

$v_f=$ final velocity of the block

$v_i=$ initial velocity of the block

When the block comes to rest it is due to the result of continuously decreasing velocity.

7 0

3 years ago

An object that floats in water weighs 20 N in air.

Licemer1 [7]

Answer:

a. Weight of Object in Water = 20 N

b. Up thrust = 20 N

c. Weight of Water Displaced = 20 N

Explanation:

The weight of the object remains same in the water as well. Because, the same force of gravity is acting there as well. Hence,

Weight of Object in Water = 20 N

Since, the object floats on the water. Therefore, according to Archimedes' principle the up thrust force acting on the object must be equal to the weight of object:

Up thrust = Weight of object

Up thrust = 20 N

From Archimedes' Principle, we know that the up thrust or the Buoyant force is equal to the weight of the water displaced by the object. therefore:

Weight of Water Displaced = Up thrust

Weight of Water Displaced = 20 N

7 0

3 years ago

Suppose a pendulum bob is swinging back and forth. If the highest point is 12.5 cm above the lowest point, what would be speed o

Colt1911 [192]

Answer:

The speed of the bob when it passes the lowest point $V = 1.57 \frac{m}{s}$