What is a common disease caused by a fungus?

Find the final velocity of a 40 kg skateboarder traveling at an initial velocity of 10 m/s that moves up a hill from a height of

slavikrds [6]

Answer:

vf = 0

Explanation:

Since the initial height hi = 0, we can rewrite the energy equation as

vf^2 = vi^2 - 2ghf = (10 m/s)^2 - 2(10 m/s^2)(5 m) = 0

Therefore, his final velocity vf is

vf = 0

3 0

3 years ago

Two thin 80.0-cm rods are oriented at right angles to each other. Each rod has one end at the origin of the coordinates, and one

kogti [31]

Answer:

The net force on the electron is given as:

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

Explanation:

Given:

charge on rod along x-axis = Q₁ = -15 x 10⁻⁶ C

charge on rod along y-axis = Q₂ = 15 x 10⁻⁶ C

distance of electron from rod 1 = r₁ = 0.4 m

distance of electron from rod 1 = r₂ = 0.4 m

charge on electron = q = -1.6 x 10⁻¹⁹ C

ε° = 8.85 x 10⁻¹² C²/Nm²

Electric force on charge due to rod 1:

F₁ = qE = 1/4πε°(qQ₁/r₁²)

F₁ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x -15 x 10⁻⁶)/0.4²

F₁ = 1.35 x 10⁻¹³ N

Negative negative repels each other so the rod will Force the electron in positive y-direction.

F₁ = 1.35 x 10⁻¹³ N j

Electric force on charge due to rod 2:

F₂ = qE = 1/4πε°(qQ₂/r₂²)

F₂ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x 15 x 10⁻⁶)/0.4²

F₂ = - 1.35 x 10⁻¹³ N

Opposite charges attract each other so the rod will force the electron in negative x-direction.

F₂ = - 1.35 x 10⁻¹³ N i

Net Force:

F = F₁ + F₂

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

4 0

3 years ago

A player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

____ [38]

At the highest point in its trajectory, the ball's acceleration is zero but its velocity is not zero.

<h3>What's the velocity of the ball at the highest point of the trajectory?</h3>

At the highest point, the ball doesn't go more high. So its vertical velocity is zero.
However, the ball moves horizontal, so its horizontal component of velocity is non - zero i.e. u×cosθ.
u= initial velocity, θ= angle of projection

<h3>What's the acceleration of the ball at the highest point of projectile?</h3>

During the whole projectile motion, the earth exerts the gravitational force with a acceleration of gravity along vertical direction.
But as there's no acceleration along vertical direction, so the acceleration along vertical direction is zero.

Thus, we can conclude that the acceleration is zero and velocity is non-zero at the highest point projectile motion.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

A- neither the ball's velocity nor its acceleration are zero.

B- the ball's acceleration points upward.

C- the ball's acceleration is zero but its velocity is not zero.

D- the ball's velocity points downward.

Learn more about the projectile motion here:

brainly.com/question/24216590

#SPJ1

7 0

2 years ago

You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c

Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

Since the block slides across the floor at constant speed, this means that it's not accelerated.
According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

$F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N (1)$

In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

$F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N (2)$

⇒ 169 N + Fn = Fg = 216 N (3)

This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
Fn = 216 N - 169 N = 47 N (4)

6 0

3 years ago

You can comfortably hold your fingers close beside a candle flame, but not very close above the flame. why? challenge (optional)

Inessa05 [86]

The candle flame releases hot gases, which directly go in upwards directions. Due to which the air near the flame of the candle is very hot and dense. The particles along with vapour move up. And since the sideways, the air is not very dense and hot, we are able to hold the candle. In anti-gravity region, there will be no density differences and also, the convection process wont occur. So, the candle quickly snuffs off.

6 0

3 years ago

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