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3 years ago

HELLLLPPPPPPPPPPPPPPPPPPPPPP PLZZZZ AND THANK YOU

Physics

1 answer:

makvit [3.9K]3 years ago

7 0

Answer:

The parents and offspring all have phenotype ww. That leaves 100% probability of short wings and 0% long wings.

Explanation:

If long wings are a dominant trait, that means short wings are recessive. If two short winged flies are crossed, each of their phenotypes are ww. Therefore, each of their offspring also have phenotype ww, or short wings.

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Which statement best describes the concept of Einstein’s equation E = mc2?

faltersainse [42]

What are the statements please

5 0

4 years ago

A projectile is fired from ground level with an initial speed of 55.6 m/s at an angle of 41.2° above the horizontal. (a) Determi

sattari [20]

Answer:

(a) t = 3.74 s

(b) H = 136.86 m

(d) ax = 0 m/s², ay = 9.8 m/s²

Explanation:

(a)

Time to reach maximum height by the projectile is given as:

t = V₀ Sinθ/g

where,

V₀ = Launching Speed = 55.6 m/s

Angle with Horizontal = θ = 41.2°

g = 9.8 m/s²

Therefore,

t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)

t = 3.74 s

(b)

Maximum height reached by projectile is:

H = V₀² Sin²θ/g

H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)

H = 136.86 m

(c)

Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:

Vₓ = V₀ₓ = V₀ Cos θ

Vₓ = (55.6 m/s)(Cos 41.2°)

Vₓ = 41.83 m/s

Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.

Vy = 0 m/s

(d)

Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.

ax = 0 m/s²

The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:

ay = 9.8 m/s²

3 0

3 years ago

A 1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the

Paul [167]

Answer:

a) 3344 N

b) 3344 N

Explanation:

This is the complete question

1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected. A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units. B. What is the magnitude of the force of the truck on the car?

Mass of the car = 1100 kg

Mass of the truck = 2200 kg

Force exerted on the ground by the car = 5000 N

The total mass in the system = 1100 + 2200 = 3300 Kg

Total force in the system = 5000 N

Recall that the force in the system = mass x acceleration

therefore,

5000 = 3300 x a

Total acceleration in the system = 5000/3300 = 1.52 m/s^2

The force on the truck individually fro the car, will be the product of this acceleration and its mass

Force on the truck = 2200 x 1.52 = 3344 N

b) Force on the car From the truck will be equal to this force but will act in the opposite direction.