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REY [17]
3 years ago
14

A body decelerates uniformly to a constant speed and after some time it accelerates uniformly Draw the shape of speed time graph

for such a motion . Label the three sections of this graph. What is the quantity which is measured by the area occupied under speed time graph ?

Physics
1 answer:
Alex73 [517]3 years ago
7 0

Answer:

here I am just giving an idea of how the graph will be like ...

In the pic..

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

:)

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An archer shoots an arrow with a mass of 45.0 grams from bow pulled
Sladkaya [172]

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

3 0
3 years ago
A square plate of edge length 9.0 cm and negligible thickness has a total charge of 6.90 10-6 C. Estimate the magnitude E of the
SVETLANKA909090 [29]

Answer:

E= 4.35*10^6 N/C

Explanation:

Let's find the area charge density of the plate

α= 6.9*10^-6/9*10^-2 = 7.7*10^-5C/m2

Now we can calculate the electric field just of the plate

E =α/2e =7.7*10^-5/2*8.85*10^-12 = 4.35*10^6 N/C

7 0
3 years ago
A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potentialenergy func
OlgaM077 [116]

Answer:

A= 148.92  m/s²

Explanation:

Given that

U(x,y) = (6.00  )x²  - (3.75  )y ³

m= 0.04 kg

Now force in the x-direction

Fx= - dU/dx

U(x,y) = (6.00  )x²  - (3.75  )y ³

dU/dx= 12 x

When x=0.4 m

dU/dx= 12 x 0.4 = 4.8

So we can say that

Fx= - 4.8 N

From Newtons law

F= m a

- 4.8 = 0.04 x a

a = -120 m/s²

Acceleration in x direction ,a = -120 m/s²

In y -direction

F= - dU/dy

U(x,y) = (6.00  )x²  - (3.75  )y ³

dU/dy = 0 - 3.75 x 3 y²

When y = 0.56 m

dU/dy = - 3.75 x 3 x 0.56 x 0.56

dU/dy = - 3.52

So we can say that force in y -direction

F= 3.52 N

F= m a'

3.52 = 0.04 x a'

a'=88.2 m/s²

acceleration in y direction is 88.2 m/s²

The resultant acceleration

A=\sqrt{a^2+a'^2}

A=\sqrt{120^2+88.2^2}

A= 148.92  m/s²

7 0
3 years ago
North America experienced all of the following during the last glacial period EXCEPT alpine glaciers covered the Rocky and Casca
Juli2301 [7.4K]

Answer:

the Hudson Bay was covered with alpine glaciers

Explanation:

During the last glacial period, large portions of North America were covered with ice. The majority of the ice was from the ice sheets that were covering Canada and the northern part of the United States, and the alpine glaciers on the mountain ranges. Hudson Bay was all frozen at this point of time. It was not covered with alpine glaciers though, instead it was covered with the ice of the extended ice sheets, with the ice cover reaching up to 2 km in thickness.

5 0
3 years ago
A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass
STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force.  We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as  588 newtons  or as 
132.3 pounds.  That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is 

                                     y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is     y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is          y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

                                                                = - (4 π² M / 15²)  sin(2π t/15)

                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of  0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of  65kg, when in reality it's only  60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

                        0.08333 G  =  0.1755 M

The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
                                       M = (0.08333 / 0.1755) (9.8) =  4.65 meters .

That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0
3 years ago
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