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3 years ago

14

In what direction does an applied force move an object

2 answers:

Dafna11 [192]3 years ago

5 0

In he same direction of the force

mr Goodwill [35]3 years ago

4 0

In the direction the force is directed towards. If the boy kicks the ball to the right, the ball will roll to the right.

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How do I go about this?

Anna71 [15]

Hi there!

(a)

Recall that:
$W = F \cdot d = Fdcos\theta$

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

$W =248(56)cos(30) = 12027.36 J$

To the nearest multiple of ten:
$W_A = \boxed{12030 J}$

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
$\boxed{W_g = 0 J}$

(c)
Similarly, the normal force is perpendicular to the displacement, so:
$\boxed{W_N = 0 J}$

(d)

Recall that the force of kinetic friction is given by:
$F_{f} =\mu_k mg$

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
$W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J$

In multiples of ten:
$\boxed{W_f = -3070 J}$

(e)
Simply add up the above values of work to find the net work.

$W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J$

Nearest multiple of ten:
$\boxed{W_{net} = 8950 J}}$

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
$F_{net} = F_{Ax} - F_f$

$W = F_{net} \cdot d = (F_{Ax} - F_f)$

$W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J$

Nearest multiple of ten:
$\boxed{W_{net} = 8950 J}$

5 0

2 years ago

Your friend wants to join the school track team, and has asked for your help to determine how fast she can run. What kind of inf

Zigmanuir [339]

Answer:

you calculate a specific type of run for example 100m and it takes 20 seconds to finish and calculate the time it takes them to finish

hope this helps

have a good day :)

Explanation:

6 0

3 years ago

The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc

qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

$R = \frac{6}{2.5} - 2.4 ohm$

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

$\frac{3R}{R+3} \times 2.5 = 6$

R= 12 ohm

2) circuit Q

$I\times (R+0.1) =V$

$R+0.1 =\frac{6}{2.5}$

R = 2.3 ohm

5 0

3 years ago

What is the volume of an object that has a density of 65g/cm3 and a mass of 130g.

lora16 [44]

Density <em>ρ</em> is mass <em>m</em> per unit volume <em>v</em>, or

<em>ρ</em> = <em>m</em> / <em>v</em>

Solving for <em>v</em> gives

<em>v</em> = <em>m</em> / <em>ρ</em>

So the given object has a volume of

<em>v</em> = (130 g) / (65 g/cm³) = 2 cm³

5 0

3 years ago

A person carries a plank of wood 2.00 m long with one hand pushing down on it at one end with a force f1 and the other hand hold

slega [8]

Answer:

F₁ = 4,120.2 N

F₂ = 3,924N

Explanation:

1) Balance of angular momentum around the end where F₁ is applied.

F₂ × 0.5m - F₁ × 0 = mass × g × 1m

⇒ F2 × 0.5 m= 20 kg × 9.81 m/s² × 1 m = 1,962 N×m

F₂ = 196.2 Nm / 0.5m = 3,924 N

2) Balance of forces

F₁ - F₂ = mg

F₁ = F₂ + mg = 3,924N + 20kg (9.81 m/s²) = 4,120.2 N

4 0

3 years ago