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3 years ago

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A farmer pulls on his obstinate mule with 250 N of force to the right. The ground exerts a reaction force to the mule’s resistan

ce of 250 N to the left. What is the net force on the mule system? What is the mule’s acceleration?

1 answer:

PtichkaEL [24]3 years ago

6 0

0 net force it is exerting an equal and opposite force against the mule. And since the net force is 0 then the acceleration would be 0 too.

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What is net force?

drek231 [11]

The answer is d my friend :)

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3 years ago

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A girl drops a stone from the top a tower 45m tall. At the same time, a boy standing at the base of the tower, projects another

Advocard [28]

Answer:

(i) The stones meet at 1.8 second

(ii) The point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.

Explanation:

(i)

First we consider the stone dropped by the girl. We have data:

Vi = Initial Velocity of Stone = 0 m/s (Since the stone was initially at rest)

t = Time Period

g = 10 m/s²

s₁ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₁ = Vi t + (0.5)gt²

s₁ = (0)(t) + (0.5)(10)t²

s₁ = 5t² ----- equation (1)

Now, we consider the stone throne vertically upward by the boy. We have data:

Vi = Initial Velocity of Stone = 25 m/s

t = Time Period

g = - 10 m/s² (negative sign due to upward motion)

s₂ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₂ = Vi t + (0.5)gt²

s₂ = (25)(t) + (0.5)(-10)t²

s₂ = 25t - 5t² ----- equation (2)

At, the point where both the stones meet, the sum of distances covered by both stones must be equal to the height of building (i.e 45 m).

s₁ + s₂ = 45

using values from equation (1) and equation (2)

5t² + 25t - 5t² = 45

25t = 45

t = 45/25

<u>t = 1.8 sec</u>

(ii)

using this value of of t in equation (2)

s₂ = (25)(1.8) - (5)(1.8)²

<u>s₂ = 28.8 m</u>

using this value of of t in equation (1)

s₁ = (5)(1.8)²

<u>s₁ = 16.2 m</u>

<u>Hence, the point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.</u>

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3 years ago

Can someone help me with accerleration

dangina [55]

Answer:

acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.

SI unit: m/s2, m·s−2, m s−2

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3 years ago

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Which expression allows you to determine the mechanical advantage of an inclined plane?

Leya [2.2K]

Answer:

(C) length / height of the plane

Explanation:

The mechanical advantage of an inclined plane can be determined using different variables. In this case, the geometry of the setup is relevant. The advantage is proportional to the length of the plane, and inversely proportional to the height: it is the ratio (length) / (height) of the plane. For example, given a desired, fixed height, a long inclined plane gives you a bigger mechanical advantage than a short inclined plane. In this example, pushing an object up the long plane will require a smaller force, than it would on the short plane.

Strictly speaking, (D) would also "allow you to determine the mechanical advantage" because you could simply invert the ratio listed under (D). However, (C) is the best, direct, answer.

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3 years ago

a 15-nC point charge is at the center of a thin spherical shell of radius 10cm, carrying -22nC of charge distributed uniformly o

Aleks [24]

Answer:

A) E = 278925.62 N/C with direction; radially out.

B) E = 43048.47 N/C with direction radially out.

C) E = -3214.29 N/C with direction radially in.

Explanation:

From Gauss' Law, the Electric field for any spherically symmetric charge or charge distribution is the same as the point charge formula. Thus;

E = kQ/r²

where;

Q is the net charge within the distance r.

We are given the charge Q = 15-nC and

spherical shell of radius 10cm

A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²

E = 278925.62 N/C

This will be radially out ,since the net charge is positive.

B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²

E = 43048.47 N/C

This will be radially out ,since the net charge is positive.

C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

Q = 15 nC - 22 nC

Q = -7 nC = -7 x 10^(-9) C

and;

E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²

E = -3214.29 N/C

This will be radially in, since the net charge is negative. You can indicate this with a negative answer.

8 0

3 years ago