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3 years ago

6

The static frictional force between a 95-kilogram object and the floor is 45 Newtons. The kinetic frictional force is only 22 Ne

wtons. What force must be exerted to accelerate the box at 0.5 meters per second to the south?

1 answer:

Lisa [10]3 years ago

6 0

Answer:

F = 69.5 [N]

Explanation:

We must remember that the friction force is defined as the product of the normal force by the coefficient of friction, and it can be calculated by the following expression.

$f=N*miu$

where:

N = normal force [N]

miu = friction coefficient

f = friction force = 22 [N]

Now we must calculate the force exerted by means of Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

$F - f = m*a$

where:

F = force exerted [N]

f = friction force [N]

m = mass = 95 [kg]

a = acceleration = 0.5 [m/s²]

Now replacing:

$F - 22 = 95*0.5\\F = 47.5 + 22\\F = 69.5 [N]$

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A single-turn square loop carries a current of 16 A . The loop is 15 cm on a side and has a mass of 3.8×10^−2kg . Initially the

DiKsa [7]

Answer:

The minimum magnetic field is 0.078 T.

Explanation:

Given that,

Current = 16 A

Side = 15 cm

Mass $m= 3.8\times10^{-2}\ kg$

Mass each segment in given square loop is

$m=\dfrac{3.8\times10^{-2}}{4}$

We need to calculate the torque due to gravity

Using formula of torque

$\tau_{g}=2mg(\dfrac{L}{2})+mgL$

$\tau_{g}=2mgL$

The torque due to magnetic field

$\tau_{B}=FL$

$\tau_{B}=BIL^2$

The equilibrium condition

$\tau_{B}=\tau_{g}$

Put the value into the formula

$BIL^2=2mgL$

$B=\dfrac{2mgL}{IL^2}$

$B=\dfrac{2mg}{IL}$

Put the value into the formula

$B=\dfrac{2\times\dfrac{3.8\times10^{-2}}{4}\times9.8}{16\times15\times10^{-2}}$

$B=0.078\ T$

Hence, The minimum magnetic field is 0.078 T.

7 0

3 years ago

In which part of the scientific method do you tell why the research is being done?

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You tell it in pupouse

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3 years ago

Read 2 more answers

21. If the Sun's rays were at 45° to a vertical pillar, how would

diamong [38]

Answer:

Let the height of the pole AB = x m. ∴ Length of shadow OB ol the pole AB = x m. Let the angle of elevation be ө, i.e. Hence, the angle of elevation of the Sun's altitude is 45°.

Explanation:

4 0

3 years ago

A van starts off 152 miles directly north from the city of Springfield. It travels due east at a speed of 25 miles per hour. Aft

erastovalidia [21]

Answer:

12.84 miles per hour

Explanation:

Given:

Vertical distance of starting point of van from Springfield (d) = 152 miles

Speed in east direction (s) = 25 mph

Distance traveled in east direction (e) = 91 miles

Let the direct distance from Springfield of the van be 'x' at any time 't'.

Now, from the question, it is clear that, the vertical distance of van is fixed at 152 miles and only the horizontal distance is changing with time.

Now, consider a right angled triangle SNE representing the given situation.

Point S represents Springfield, N represents the starting point of van and E represents the position of van at any time 't'.

SN = d = 152 miles (fixed)

Now, using the pythagorean theorem, we have:

$SE^2=SN^2+NE^2\\\\x^2=d^2+e^2\\\\x^2=(152)^2+e^2----(1)$

Now, differentiating both sides with respect to time 't', we get:

$2x\frac{dx}{dt}=0+2e\frac{de}{dt}\\\\\frac{dx}{dt}=\frac{e}{x}\frac{de}{dt}$

Now, we are given speed as 25 mph. So, $\frac{de}{dt}=25\ mph$

Also, when $e=91\ mi$ , we can find 'x' using equation (1). This gives,

$x^2=23104+(91)^2\\\\x=\sqrt{31385}=177.16\ mi$

Now, plug in the values of 'e' and 'x' and solve for $\frac{dx}{dt}$ . This gives,

$\frac{dx}{dt}=\frac{91}{177.16}\times 25\\\\\frac{dx}{dt}=12.84\ mph$

Therefore, the distance between the van and Springfield is changing at a rate of 12.84 miles per hour

6 0

3 years ago

1pt A cannon fires a 5-kg ball horizontally from a

Klio2033 [76]

Answer: Both cannonballs will hit the ground at the same time.

Explanation:

Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.

then the acceleration equation is only on the vertical axis, and can be written as:

a(t) = -(9.8 m/s^2)

Now, to get the vertical velocity equation, we need to integrate over time.

v(t) = -(9.8 m/s^2)*t + v0

Where v0 is the initial velocity of the object in the vertical axis.

if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s

and:

v(t) = -(9.8 m/s^2)*t

Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)

And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)

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3 years ago