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shusha [124]
3 years ago
7

An object moves in a circle at a constant speed of 1.0 m/s. The radius of the circle is 1.0 m. If a force of 1.0 N acts toward t

he center of the circle, how much work does this force do as the object moves? 1. W = F (2 π r) for a circle radius of r. 2. W = F (π r2 ) for a circle radius of r. 3. Zero 4. W = m v2 r for a circle radius of r. 5. None of these
Physics
1 answer:
Vlad1618 [11]3 years ago
4 0

Answer:5

Explanation:

Given

speed of object v=1\ m/s

radius of circle r=1\ m

Force towards the center F=1\ N

Work done is given by the dot product of Force and displacement

and we know know displacement of the object is along the circle which is perpendicular to the force acting therefore Work done will be zero

W=F\cdot s\cos 90

W=0

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Out put work
explanation:
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3 years ago
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a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring force constant with
icang [17]

Complete question:

a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring with force constant of  955 N/m. The block comes to rest after compressing the spring a distance of 4.6 cm. Find the initial speed (in m/s) of the block.

Answer:

The initial speed of the block is 1.422 m/s

Explanation:

Given;

mass of the block, m = 2.0 kg

force constant of the spring, K = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

Apply Hook's law to determine applied force on the spring;

F = Kx

F = (955 N/m)(0.046 m)

F = 43.93 N

Apply Newton's 2nd law to determine the magnitude of deceleration of the block when it encounters the spring;

F = ma

a = F / m

a = 43.93 / 2

a = 21.965 m/s²

Apply kinematic equation to determine the initial speed of the block;

v² = u² + 2ax

where;

v is the final speed of the block = 0

u is the initial speed of the block

x is the distance traveled by the block = compression of the spring

a is the block deceleration = -21.965 m/s²

0 = u² + 2(-21.965 )(0.046)

0 = u²  - 2.021

u² =  2.021

u = √2.021

u = 1.422 m/s

Therefore, the initial speed of the block is 1.422 m/s

8 0
3 years ago
A current exists whenever electric charges move. If ΔQ is the net charge that passes through a surface during a time period Δt,
jeka57 [31]

Answer:

It represents the change in charge Q from time t = a to t = b

Explanation:

As given in the question the current is defined as the derivative of charge.

                                  I(t) = dQ(t)/dt ..... (i)

But if we take the inegral of the equation (i) for the time interval  from t=a to

t =b we get

                                   Q =∫_a^b▒〖I(t)  〗 dt

which shows the change in charge Q from time t = a to t = b. Form here we can say that, change in charge is defiend as the integral of current for specific interval of time.

5 0
3 years ago
On March 21, a stick casts the following shadow. What is the most likely time of day?
jenyasd209 [6]

Answer:

9:00 AM

Explanation:

I took the test and that was the answer

5 0
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Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th
brilliants [131]

From the case we know that:

  1. The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
  2. The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
  3. The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

#SPJ4

8 0
1 year ago
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