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Degger [83]
1 year ago
13

Force Problem: Four concurrent forces are applied to a 5 Kg mass: 50 N North, 25 N East, 35 N South, and 35 N West.

Physics
1 answer:
Kaylis [27]1 year ago
6 0

A) Net horizontal force:
Fx= -35 + 25 + -10N

Net vertical force
Fy= 50 - 35 = 15N

Net force
F^2 = fX^2 +Fy^2
F= 18N

B.) Acceleration of system
a=18/5=3.6m/s^2

C). Equilibriant force must be equal to the net force above
F1= 18N

D.) Acceleration becomes half
a1= a/2 =1.8m/s^2

E.) Equilibrant force also doubles
F" = 2F' = 36N

To learn more about Force, click on the link
brainly.com/question/12785175

#SPJ13

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A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The to
SpyIntel [72]
This is a problem of conservation of momentum

Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s

A) man throws the rock forward

=>

rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man

sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?

Conservation of momentum:
momentum before throw = momentum after throw

46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2

=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s

B) man throws the rock backward

this changes the sign of the velocity, v2 = -14.5 m/s

 46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2

v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s


3 0
3 years ago
Two particles, each with charge Q, and a third charge q, are placed at the vertices of an equilateral triangle as shown. The tot
Gelneren [198K]

Answer:

<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>

Explanation:

The image is shown below.

The force on the particle with charge q due to each charge Q = \frac{kQq}{r^{2} }

we designate this force as N

Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.

Resolving the forces on the particle, we have

for the x-component

N_{x} = N cosine 60° + (-N cosine 60°) = 0

for the y-component

N_{y} = -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N

The above indicates that there is no resultant force in the x-axis, since it is equal to zero (N_{x} = 0).

The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.

<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>

5 0
4 years ago
A 12.7 L box is filled with 6.0 moles of Argon gas and cooled to a temperature of 210 K. What is the pressure?
sp2606 [1]

Answer:

Gases are easily compressed. We can see evidence of this in Table 1 in Thermal Expansion of Solids and Liquids, where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same β. This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.

Explanation:

3 0
3 years ago
On the Moon, the acceleration due to the effect of gravity is only about 1/6 of that on Earth. An astronaut whose weight on Eart
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A = 94.22 Newtons

b = 58.16 kg

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8 0
4 years ago
6. A satellite is orbiting Earth just above its surface. The centripetal force making the satellite follow a circular trajectory
Svetllana [295]

Answer:

Engular velocity: w=1,24*10^{-3}/s

Linear velocity: V=7905 m/s

The time it takes:

t=5060s=84min

Explanation:

The magnitude of the centripetal acceleration can be related to the angular velocity and radius as:

(1)a=r*w^{2}

Solving for w:

(2)w=\sqrt{\frac{a}{r} }

Replacing a=9,8m/s2 and r=6,375,000m:

(3)w=\sqrt{\frac{9,8m/s^{2}}{6375000m} }=1,24*10^{-3}/s

And the angular velocity relates to the linear velocity:

V=w*r=1,24*10^{-3}/s*6375000m=7905 m/s

The perimeter of the orbit is:

P=\pi *2*r=\pi *2*6375000m=40.05*10^{6}m

The time it takes:

t=\frac{P}{V} =\frac{40.05*10^{6}m}{7905 m/s}=5060s=84min

5 0
3 years ago
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