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Troyanec [42]
3 years ago
9

A massive block is being pulled along a horizontal frictionless surface by a constant horizontal force. The block must be ______

____. View Available Hint(s) A massive block is being pulled along a horizontal frictionless surface by a constant horizontal force. The block must be __________. continuously changing direction moving at constant velocity moving with a constant nonzero acceleration moving with continuously increasing acceleration
Physics
1 answer:
snow_tiger [21]3 years ago
4 0

Answer:

The body must be moving with a constant non zero acceleration.

Explanation:

Force produces acceleration on any mass it is applied on. The acceleration produced depends on the magnitude and direction of the force. For this block being dragged by a constant horizontal force, The body will undergo a constant non-zero acceleration that will steadily increase its velocity along the direction of the force.

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A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor
Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force F_{1}=6\ N

Second force F_{2}=10\ N

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

\tau=Force\times lever\ arm

So, Here,

\tau_{2}=5 \tau_{1}

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

F_{2}\times d_{2}=5\times F_{1}\times r_{1}

r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}

Where, F_{1}=First force

F_{1}=First force

F_{2}=Second force

r_{1}= distance

Put the value into the formula

r_{2}=\dfrac{5\times6\times0.2}{10}

r_{2}=0.6\ m

Hence, The moment arm is 0.6 m

6 0
3 years ago
How much electrical is use by a 350 W television that is operating for 25 minutes
Komok [63]

350Js^¹- *25*60s=525000J

7 0
3 years ago
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Explain how the earth is round and not flat
frosja888 [35]
It is a conglomeration of rock that collected together over time, and it rotating in a spinning ball that is orbiting a sun that is in the middle of our solar system. 

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3 years ago
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An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t
viktelen [127]

Answer:

a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

          F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0
3 years ago
The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for
pshichka [43]

Answer:

d = 69 .57 meter

Explanation:

First case

Speed of car ( v )  = 20.5 mi/h  = 9.164  M/S

distance ( d ) = 11.6 meter                                       ( m = mass of the car )

Work done = 0.5 m v²  = 0.5 * 9.164² * m J  = 41.99 m J

Force = ( workdone /distance ) = ( 41.99 m / 11.6 )   =  3.619 m N

Second case

v = 50.2 mi/h = 22.44135 m/s

d = ?

Work done = 0.5 * 22.44² * m J = 251.7768 * m J

Since the braking force remains the same .

3.619 m = ( 251.7768 m / d )

d = 69 .57 meter

7 0
3 years ago
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