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2 years ago

Find the work done in moving a particle from p to q if the magnitude and direction

Physics

1 answer:

exis [7]2 years ago

3 0

<span>Work=32unit</span><span>...hopes thats helps</span>

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Gantt charts are A. widely used network techniques. B. not widely used. C. planning charts used to schedule resources and alloca

kotykmax [81]

Answer:

planning charts used to schedule resources and allocate time is the correct answer

Explanation:

Gantt Charts are a project planning tool used to track the projects from the start to endpoint.

Gantt charts are used for the preparation and planning of the projects.

By using Gantt charts we can assess how much time a project would take to complete and we can also determine resources required with the help of Gantt charts.

6 0

3 years ago

Help me pls I WILL GIVE BRAINIEST I DONT CARE PLS HELP

PilotLPTM [1.2K]

Answer:

A) was reusable

Explanation:

Check this website out for more information about the space shuttle: https://www.nasa.gov/audience/forstudents/k-4/stories/nasa-knows/what-is-the-space-shuttle-k4.html

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2 years ago

Read 2 more answers

A golf club hits a stationary 0.05kg golf ball with and average force of 5.0 x 10^3 newtons accelerating the ball at 44 meters p

maxonik [38]

Answer: The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

Explanation:

Force applied on the golf ball = $5.0\times 10^3 N$

Mass of the ball = 0.05 kg

Velocity with which ball is accelerating = 44 m/s

Time period over which forece applied = t

$f=ma=\frac{m\times v}{t}$

$t=\frac{0.05 kg\times 44m/s}{5.0\times 10^3 N}=4.4\times 10^{-4} seconds$

$Impulse=(force)\times (time)=f\times t = 5.0\times 10^3\times 4.4\times 10^{-4} seconds=2.2$ Newton seconds

The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

7 0

2 years ago

Can someone help me

SashulF [63]

That is b. hope this helps cx

8 0

2 years ago

An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields a

Lapatulllka [165]

Answer:

The speed of the electron is $2.55\times 10^3\ m/s$ .

Explanation:

Given that,

The magnitude of electric field, $E=1.32\ kV/m=1.32\times 10^3\ V/m$

The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then, the magnitude of electric field and magnetic field is balanced such that :

$evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s$

$v=2.55\times 10^3\ m/s$

So, the speed of the electron is $2.55\times 10^3\ m/s$ . Hence, this is the required solution.

3 0

3 years ago