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Alex17521 [72]
3 years ago
12

An initially stationary 2.7 kg object accelerates horizontally and uniformly to a speed of 13 m/s in 4.0 s. (a) In that 4.0 s in

terval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?
Physics
1 answer:
Ronch [10]3 years ago
8 0

Explanation:

A.

Given:

V = 13 m/s

t = 4 s

Constant acceleration, a= (V-Vi)/t

= 13/4

= 3.25 m/s^2

F = mass * acceleration

= 2.7 * 3.25

= 8.775 N.

Using equations of motion,

distance,S = (13 * 4) - (1/2)(3.25)(4^2)

= 26 m

Workdone, W = force * distance

= 8.775 * 26

= 228.15 J

B.

Instantaneous power, P = Force *Velocity

= 8.775 * 13

= 114. 075 W

C.

t = 2 s,

Constant acceleration, a= (V-Vi)/t

= 13/2

= 6.5 m/s^2

Force = mass * acceleration

= 2.7 * 6.5

= 17.55 N

Instantaneous power, P = Force *Velocity

= 17.55 * 13

= 228.15 W.

= 114. 075 W.

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arlik [135]

Answer:

Part a)

a = 1260.3 m/s^2

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

v_f^2 - v_i^2 = 2 a d

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v_1 = 8.86 m/s

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

v_f^2 - v_i^2 = 2 a d

0 - v_2^2 = 2(-9.81)(2.00)

v_2 = 6.26 m/s

Part a)

Average acceleration is given as

a = \frac{v_f - v_i}{\Delta t}

a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}

a = 1260.35 m/s^2

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0
3 years ago
What do the top of a zip line and the top of a bungee cord have in common
LiRa [457]

Answer:

They are both placed at high vantage points for an optimal experience.

Explanation:

Gravity works in your favor when participating in bungee jumping as well as ziplining

6 0
2 years ago
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A hungry rat is placed in a maze. It walks the following path to find a piece of cheese. 4.0m N, 7.5 m E, 6.8 m S, 3.7 m E, 3.6
storchak [24]

Answer:

Explanation:

We shall take the help of vector form of displacement . Taking east as i and north as j

4.0m N = 4 j

7.5 m E = 7.5 i

6.8 m S = - 6.8 j

3.7 m E, = 3.7 i

3.6 m S = - 3.6 j

5.3 m W = - 5.3 i

3.7 m N, = 3.7 j

5.6 m W = - 5.6 i

4.4 m S = - 4.4 j

4.9 m W = -  4.9 i

Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i

= -4.6 i -7.1 j

magnitude of displacement = \sqrt{(4.6^2+7.1^2)}

= 8.46 m

Direction

Tanθ = 7.1/ 4.6

θ = 57⁰ south of west .

distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9

= 49.5 m

6 0
3 years ago
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Halley is standing outside on a cloudy day. When she hears thunder, she goes back inside so that she doesn't get caught in
jarptica [38.1K]

Answer:

Inference

Explanation:

An inference involves the application of logic to progress from a premise to a conclusion or logical consequence on the basis of the evidence or known fact. Inference is a process of thought that be divided into a deduction and an induction aspect.

In the given question Halley, by standing outside was able to deduce the sound of thunder she is then able by inductive reasoning from the fact that storms are usually preceded by and accompany lightening, conclude that there is a storm coming.

6 0
3 years ago
A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° below the
Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, (\theta) = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ F_x=Fcos(\theta)          and             ⇒ F_y=Fsin (\theta)

⇒ F_x=15cos(15)                           ⇒ F_y=15sin (15)

⇒ Applying concept of  forces.

⇒ \sum F_x=F_n_e_t =F-f

⇒ F_n_e_t =F-f

⇒ ma =F-f       <em>  ...Newtons second law Fnet = ma</em>

⇒ a =\frac{F-f}{m}              

⇒ Plugging the values.

⇒ a =\frac{15cos(15)-0}{40}     <em>...f is the friction which is zero here.</em>

⇒ a =\frac{14.48}{40}

⇒ a=0.362\ ms^-^2

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0
3 years ago
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