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3 years ago

Help pls will mark right answer

Physics

1 answer:

german3 years ago

5 0

Answer:

Less

Explanation:

Since acceleration is inversely propotional to mass keeping the force constant, hence more mass will accelerate less and vice versa.

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2.0 kg mass is lifted4.0m above the ground find the gravitational energy

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E = mgh
= (2)(9.8)(4)
= 78.4 Joules

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3 years ago

What do bats, dolphin, and whale use to determine their location?

Flura [38]

C. Echolocation

Echolocating animals emit calls out to the environment and listen to the echoes of those calls that return from various objects near them. They use these echoes to locate and identify the objects.

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3 years ago

Joe and Bill throw identical balls vertically upward. Joe throws his ball with an initial speed twice as high as Bill. If there

exis [7]

Answer:

the maximum height of Joe's ball will be 4 times Bill's ball.

Explanation:

let bill's velocity be v then Joe's velocity is 2v.

and initial velocities of both bill and Joe are 0

for Bill

$v^2-u^2= 2gh$

$h=\frac{v^2}{2gh}$

for Joe

$(2v)^2-u^2= 2gh'$

$h'=\frac{4v^2}{2g}$

thus we can write that

h'=4h

the maximum height of Joe's ball will be 4 times Bill's ball.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Chuge%20%5Cfrak%7B%E0%BC%86%20%20%5C%3A%20%20%5C%3A%20question%20%5C%3A%20%20%5C%3A%20%E0%BC

My name is Ann [436]

The crate is moving at constant velocity when the forces acting on it are

balanced.

The value of the force <em>F</em> required to pull the crate with constant velocity is; $\underline{F = \sqrt{2} \cdot \mu \cdot m\cdot g}$

Reasons:

Mass of the crate = m

Cross section of through = Right angled

Orientation (inclination) of the through to horizontal = 45°

Coefficient of kinetic friction = μ

Required:

The value of the required to pull the crate along the through at constant

velocity.

Solution:

When the through is moving at constant velocity, we have;

Friction force acting on crate = Force pulling the crate

Friction force = Normal reaction × Coefficient of kinetic friction

Normal reaction on an inclined plane = $\mathbf{F_N}$

Each side of the through gives a normal reaction.

The vertical component of the normal reaction on each side of the through

is therefore;

$F_N$ ·j = $\mathbf{F_N}$ × sin(θ)

The sum of the vertical component = $F_N$ ·j + $F_N$ ·j = 2· $F_N$ ·j = 2· $F_N$ ×sin(θ)

The sum of the vertical component of the normal reactions = The weight of the crate

Therefore;

2· $F_N$ ×sin(θ) = m·g

θ = 45°

Therefore;

2· $F_N$ ×sin(45°) = m·g

$\displaystyle sin(45^{\circ}) = \mathbf{\frac{\sqrt{2} }{2}}$

Therefore;

$\displaystyle 2 \cdot F_N \cdot sin(45^{\circ}) = 2 \cdot F_N \times \frac{\sqrt{2} }{2} = \sqrt{2} \cdot F_N$

$\displaystyle F_N = \mathbf{ \frac{m \cdot g}{\sqrt{2} }}$

Which gives;

$\displaystyle Force \ required, \ F = Sum \of \ friction \ forces \ = 2 \times F_N \times \mu = \mathbf{ 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu}$

$\displaystyle Force \ required, \ F = 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu = \mathbf{ \sqrt{2} \cdot \mu \cdot m \cdot g}$

Force required to pull the crate at constant velocity, <u>F = √2·μ·m·g</u>

Learn more about force of friction here:

brainly.com/question/6561298

8 0

2 years ago

Read 2 more answers

You nose out another runner to win the 100.000m dash. If your total time for the race was 11.8 seconds and you aced out the othe

daser333 [38]

You completed 100.000 m in 11.800 second

you aced out the other runner by 0.001 seconds

time taken by other runner to finish = 11.800 + 0.001 = 11.801s

speed of other runner = 100/11.801 m/s

distance covered by him in 11.8 s = 100/11.801 × 11.8 = 99.992 m

you won by = 100 - 99.992 = 0.008 meter

3 0

4 years ago