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3 years ago

The forces exerted on an object are shown. (3 points)

Physics

1 answer:

Alex_Xolod [135]3 years ago

7 0

Answer:

<em>F equals 3 N and the object remains stationary</em>. (second option in the list)

Explanation:

For sure to cancel acting forces, F must be 3N pointing up. But with regards to the object stationary or not, the question is tricky. We could have a ZERO net force applied, and the object moving at constant speed, which could still verify Newton's Laws. But considering the first answer option that refers to vertical motion upward where the object could be gaining potential energy, the most accurate response is that the force F has to be 3 N pointing up to make the object in equilibrium, and no motion in the vertical axis.

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The chart shows data for a moving object.

Ipatiy [6.2K]

Answer:

number 2

Explanation:

8 0

3 years ago

What can alter the motion of an object

Liula [17]

Force can alter its direction,slow or stop it you could say it can change its velocity

3 0

3 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

A fan cart initially has an acceleration of 1.6m/s/s when it's fan is directed straight backwards. If you rotate the fan by 45°,

Sholpan [36]

Answer:

A Fan Cart Initially Has An Acceleration Of 1.6m/s/s When It's Fan Is Directed Straight Backwards. If You Rotate The Fan By 45o, By What Percentage Do You Expect The Fan Cart's Thrust To Decrease? (Answer Should Be In Units Of 96)

a. 45%

b. 29%

c. 71%

d. 50%

The correct answer is d.

d. 50%

Explanation:

Fan cart acceleration = 1.6 m/s²

Thrust = 0.25×π×D²×ρ×v×Δv

where Δv = acceleration component and all factors remaining cconstant, when the fan is rotated by 45 ° the diameter changes to D₂ = sin 45 ×D

or 0.707×D. The thrust becomes 0.25×π×(0.707×D)²×ρ×v×Δv

=0.25×π×0.5×D²×ρ×v×Δv or 0.5(0.25×π×D²×ρ×v×Δv)

That is the thrust reduces by 50 %

3 0

3 years ago

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.550 m/s. The to

Arisa [49]

Answer: 0.5 m/s

Explanation:

Given

Speed of the sled, v = 0.55 m/s

Total mass, m = 96.5 kg

Mass of the rock, m1 = 0.3 kg

Speed of the rock, v1 = 17.5 m/s

To solve this, we would use the law of conservation of momentum

Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns

When the man throws the rock forward

rock:

m1 = 0.300 kg

V1 = 17.5 m/s, in the same direction of the sled with the man

m2 = 96.5 kg - 0.300 kg = 96.2 kg

v2 = ?

Law of conservation of momentum states that the momentum is equal before and after the throw.

momentum before throw = momentum after throw

53.08 = 0.300 * 17.5 + 96.2 * v2

53.08 = 5.25 + 96.2 * v2

v2 = [53.08 - 5.25 ] / 96.2

v2 = 47.83 / 96.2

v2 = 0.497 ~= 0.50 m/s

3 0

3 years ago