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3 years ago

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The velocity-time and acceleration-time graphs of a particle are given here. Its position-time graph may be given as: PLZ answer

fast!!!!! I will mark best as BRAINLEST!!!!!!! PLZ!!! TEST IN 2O MINS

1 answer:

artcher [175]3 years ago

8 0

Answer:

I would say the answer is B

Explanation:

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A 4.0kg block is sliding with a constant velocity of 3.0m/s on a frictionless table that is 0.5m high. If all of the block’s ene

MArishka [77]

Answer:

Velocity = 4.33[m/s]

Explanation:

The total energy or mechanical energy is the sum of the potential energy plus the kinetic energy, as it is known the velocity and the height, we can determine the total energy.

$E_{M}=E_{p} + E_{k} \\E_{p} = potential energy [J]\\E_{k} = kinetic energy [J]\\where:\\E_{p} =m*g*h\\E_{p} = 4*9.81*0.5=19.62[J]\\E_{k}=\frac{1}{2} *m*v^{2} \\E_{k}=\frac{1}{2} *4*(3)^{2} \\E_{k}=18[J]\\Therefore\\E_{M} =18+19.62\\E_{M}=37.62[J]$

All this energy will become kinetic energy and we can find the velocity.

$37.62=\frac{1}{2} *m*v^{2} \\v=\sqrt{\frac{37.62*2}{4} } \\v=4.33[m/s]$

8 0

3 years ago

A motorboat traveling on a straight course slows

never [62]

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

65 km/h → 18.0556 m/s
35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

v² = v₀² + 2aΔx

Substitute the known values into the equation.

(9.72222)² = (18.0556)² + 2a(45)
94.52156173 = 326.0046914 + 90a
-231.4831296 = 90a
a = -2.572

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

3 0

3 years ago

The formation of a cell plate is a characteristic of

RUDIKE [14]

The formation of a cell plate is a characteristic of cytokinesis in terrestrial plants.

4 0

3 years ago

A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upwar

sesenic [268]

Answer:

6.0 m below the top of the cliff

Explanation:

We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation

$v^2-u^2 = 2gd$

where

u = 0 (it starts from rest)

g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)

h = 24 m is the distance covered

Solving for h,

$v=\sqrt{2gh}=\sqrt{2(9.8)(24)}=21.7 m/s$

So the ball thrown upward is launched with this initial velocity:

u = 21.7 m/s

From now on, we take instead upward as positive direction.

The vertical position of the ball dropped from the cliff at time t is

$y_1 = h - \frac{1}{2}gt^2$

While the vertical position of the ball thrown upward is

$y_2 = ut - \frac{1}{2}gt^2$

The two balls meet when

$y_1 = y_2\\h-\frac{1}{2}gt^2 = ut - \frac{1}{2}gt^2 \\h = ut \rightarrow t = \frac{h}{u}=\frac{24}{21.7}=1.11 s$

So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is

$y_1 = h -\frac{1}{2}gt^2 = 24-\frac{1}{2}(9.8)(1.11)^2=18.0 m$

So the distance below the top of the cliff is

$d=24.0 - 18.0 = 6.0 m$

4 0

3 years ago

suppose you increase your walking speed from 5m/s to 14m/s in a period of 1 . what is your acceleration?

Ne4ueva [31]

Acceleration = (change in speed) / (time for the change.

change in speed = (ending speed) - (starting speed) = 9 m/s.

Acceleration = (9 m/s) / (period of 1) .

We don't know the units of the 'period of 1'.
If it means '1 second', then the acceleration is 9 m/s² .

7 0

3 years ago