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3 years ago

Please help will give brainliest !

Physics

1 answer:

sergij07 [2.7K]3 years ago

5 0

Answer:

Explanation:

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A particle moves along a straight path through displacement while force acts on it. (Other forces also act on the particle.) Wha

allsm [11]

a) c = 1.85

b) c = 0.8

c) c = 2.33

Explanation:

The displacement of the particle is given by

$d=2.2i+cj$

While the force applied on the particle is

$F=3.2i-3.8 j$

So we have a problem in 2-dimensions.

The work done on the particle is given by the scalar product between force and displacement:

$W=F\cdot d$ (1)

Here the work done on the particle is zero, so

W = 0

Therefore from eq(1) we find:

$0=(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=7.04\\c=\frac{7.04}{3.8}=1.85$

In this problem, the work done on the particle is

$W=4.0 J$

The force and displacement are still

$d=2.2i+cj$ (displacement)

$F=3.2i-3.8 j$ (force)

Therefore, by calculting the scalar product between force and displacement and equating it to the work done (4.0 J), we find:

$W=F\cdot d$

$4.0 =(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=3.04\\c=\frac{3.04}{3.8}=0.8$

In this problem instead, the work done on the particle is negative:

$W=-1.8 J$

As before, the force and displacement are

$d=2.2i+cj$ (displacement)

$F=3.2i-3.8 j$ (force)

And so again, we calculate the scalar product between force and displacement and we equate it to the work done on the particle, -1.8 J.

Doing so, we find:

$W=F\cdot d$

$-1.8=(3.2i-3.8j)\cdot (2.2i+c)=7.04-3.8c\\3.8c=8.84\\c=\frac{8.84}{3.8}=2.33$

7 0

3 years ago

Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial veloc

svet-max [94.6K]

Incomplete question as time is missing.I have assumed some times here.The complete question is here

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Explanation:

Given data

Vi=10 m/s

S=70 m

(a) t₁=0.5 s

(b) t₂=1 s

(d) t₄=2 s

(e) t₅=2.5 s

To find

Displacement S from t₁ to t₅

Velocity V from t₁ to t₅

Solution

According to kinematic equation of motion and given information conclude that v is given by

$v=v_{i}+gt\\$

Also get the equation of displacement

$S=v_{i}t+(1/2)gt^{2}$

These two formula are used to find velocity as well as displacement for time t₁ to t₅

For t₁=0.5 s

$v_{1}=v_{i}+gt\\v_{1}=(10m/s)+(9.8m/s^{2} ) (0.5s)\\v_{1}=14.9m/s\\ And\\S_{1} =v_{i}t+(1/2)gt^{2}\\ S_{1}=(10m/s)(0.5s)+(1/2)(9.8m/s^{2} )(0.5s)^{2} \\S_{1}=6.225m$

For t₂

$v_{2}=v_{i}+gt\\v_{2}=(10m/s)+(9.8m/s^{2} ) (1s)\\v_{2}=19.8m/s\\ And\\S_{2} =v_{i}t+(1/2)gt^{2}\\ S_{2}=(10m/s)(1s)+(1/2)(9.8m/s^{2} )(1s)^{2} \\S_{2}=14.9m$

For t₃

$v_{3}=v_{i}+gt\\v_{3}=(10m/s)+(9.8m/s^{2} ) (1.5s)\\v_{3}=24.7m/s\\ And\\S_{3} =v_{i}t+(1/2)gt^{2}\\ S_{3}=(10m/s)(1.5s)+(1/2)(9.8m/s^{2} )(1.5s)^{2} \\S_{3}=26.025m$

For t₄

$v_{4}=v_{i}+gt\\v_{4}=(10m/s)+(9.8m/s^{2} ) (2s)\\v_{4}=29.6m/s\\ And\\S_{4} =v_{i}t+(1/2)gt^{2}\\ S_{4}=(10m/s)(2s)+(1/2)(9.8m/s^{2} )(2s)^{2} \\S_{4}=39.6m$

For t₅

$v_{5}=v_{i}+gt\\v_{5}=(10m/s)+(9.8m/s^{2} ) (2.5s)\\v_{5}=34.5m/s\\ And\\S_{5} =v_{i}t+(1/2)gt^{2}\\ S_{5}=(10m/s)(2.5s)+(1/2)(9.8m/s^{2} )(2.5s)^{2} \\S_{5}=55.625m$

4 0

4 years ago

how high a hill would a 75 kg hiker has to climb to increase her gravitational potential energy by 10000 j

ikadub [295]

Your answer would be 13.6 m

7 0

3 years ago

Read 2 more answers

If a spring is stretched 4m from its starting length when 20n of force is applied, then how much work (in joules) is done by the

lys-0071 [83]

ANSWER:

250 J

STEP-BY-STEP EXPLANATION:

F = 20N is required to stretch the spring by 4 meters

We know that the force is equal to:

$F=k\cdot x$

We solve for k (spring constant):

$k=\frac{F}{x}=\frac{20}{4}=5\text{ N/m}$

The work done in stretching the spring is given by the following equation (in this case the stretch is 10 meters:

$\begin{gathered} W=\frac{1}{2}k\cdot x^2 \\ \text{ Replacing} \\ W=\frac{1}{2}\cdot5\cdot10^2 \\ W=250\text{ J} \end{gathered}$

The work required is 250 joules.

5 0

2 years ago

How is kinetic energy related to potential energy?

11111nata11111 [884]

Answer:

Potential energy is energy stored in an object due to its position or arrangement. Kinetic energy is energy of an object due to its movement - its motion. ... Potential energy can be converted into kinetic energy, and kinetic energy can be converted into potential energy.

Potential energy (PE) is stored energy due to position or state.

Kinetic energy (KE) is energy of motion.

<u>Law of Conservation of Energy</u> states that <em>energy cannot be destroyed but can only be transformed form one form into another.</em> KE and PE are interconvertible and hence do not contradict the Law of Conservation of Energy.

PE + KE = Total Energy

4 0

3 years ago