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Sergeeva-Olga [200]
3 years ago
11

A potential energy function for a system in which a two-dimensional force acts is of the form U = 3x6y − 5x. Find the force that

acts at the point (x, y). (Use the following as necessary: x and y.)
Physics
1 answer:
liq [111]3 years ago
7 0

Answer:

Force, F=((-18x^5y+5)i+(-3x^6)j)\ N

Explanation:

Given that,

A potential energy function for a system in which a two-dimensional force acts is of the form of :

U=3x^6y-5x

We need to find the force that acts at the point (x, y). The force in 2 dimensional with components is given by :

F=(\dfrac{-dU}{dx},\dfrac{-dU}{dy})\\\\F=(\dfrac{-d(3x^6y-5x)}{dx},\dfrac{d(3x^6y-5x)}{dy})\\\\F=(-(18x^5y-5)),(-(3x^6))\\\\F=((-18x^5y+5)i+(-3x^6)j)\ N

So, the force acting at the point (x,y) is ((-18x^5y+5)i+(-3x^6)j)\ N. Hence, this is the required solution.

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Margarita [4]

Answer:

a)H=84.87 m

b)R=565.27 m

Explanation:

Given that

Speed of ball on earth

U=41.2 m/s

θ=31°

We know that

Maximum height in projectile motion given as

h=\dfrac{U^2sin^2\theta }{2g}

h=\dfrac{41.2^2sin^231}{2\times 9.81}

h=22.94 m

Range in projectile motion given as

r=\dfrac{U^2sin2\theta }{g}

r=\dfrac{41.2^2sin62 }{g}

r=152.77 m

Given that on distance planet ball moves 3.7 times far more as on earth.

So on the distance planet

The maximum height ,H=3.7 h

H= 3.7 x 22.94

H=84.87 m

The range on distance planet

R=3.7 r

R=3.7 x 152.77 m

R=565.27 m

7 0
3 years ago
Shelley is in an elevator that is traveling downward and slowing down at a rate of
liraira [26]

Answer:

N = 648.55[N]

Explanation:

To solve this problem we must use 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 = m*a

where:

∑F =  Forces applied [N]

m = mass = 73.2 [kg]

a = acceleration = 0.950 [m/s²]

Let's assume the direction of the upward forces as positive, just as if the movement of the box is upward the acceleration will be positive.

By performing a summation of forces on the vertical axis we obtain all the required forces and other magnitudes to be determined.

-m*g + N = -m*a\\

where:

g = gravity acceleration = 9.81 [m/s²]

N = normal force (or weight) measured by the scale = 83.4 [N]

Now replacing:

-(73.2*9.81)+N=-73.2*0.950\\-718.092+N=-69.54\\N = -69.54+718.092\\N = 648.55[N]

The acceleration has a negative sign, this means that the elevator is descending at that very moment.

8 0
3 years ago
A bin is given a push across a horizontal surface. The bin has a mass m, the push gives it an initial speed of 1.60 m/s, and the
emmasim [6.3K]

Answer:

The bin moves 0.87 m before it stops.

Explanation:

If we analyze the situation and apply the law of conservation of energy to this case, we get:

Energy Dissipated through Friction = Change in Kinetic Energy of Bin (Loss)

F d = (0.5)(m)(Vi² - Vf²)

where,

F = Frictional Force = μR    

but, R = Normal Reaction = Weight of Bin = mg

Therefore, F = μmg

Hence, the equation becomes:

μmg d = (0.5)(m)(Vi² - Vf²)

μg d = (0.5)(Vi² - Vf²)

d = (0.5)(Vi² - Vf²)/μg

where,

Vf = Final Velocity = 0 m/s (Since, bin finally stops)

Vi = Initial Velocity = 1.6 m/s

μ = coefficient of kinetic friction = 0.15

g = 9.8 m/s²

d = distance moved by bin before coming to stop = ?

Therefore,

d = (0.5)[(1.6 m/s)² - (0 m/s)²]/(0.15)(9.8 m/s²)

<u>d = 0.87 m</u>

5 0
3 years ago
According to the directions on a can of frozen orange juice concentrate, 1 can of concentrate is to be mixed with 3 cans of wate
WARRIOR [948]

Answer:

25

Explanation:

Given:

1 can of concentrate requires 3 cans of water

Now,

Total ounces in 200 6-ounce cans = 1200 ounces

also,

for 1 can of concentrate requires 3 cans of water

thus,

for 12 ounces can water can required = 3 × 12 ounces = 36 ounces of cans

Thus,

total ounce of juice per can = 12 + 36 = 48 ounces per can

therefore,

the number of 12-ounce cans required  are = \frac{\textup{Ttoal ounces}}{\textup{Ounces per can}}

or

= \frac{\textup{1200}}{\textup{48}}

or

the number of 12-ounce cans required  are = 25

8 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.20×10^6 N, one at an angle 14.0∘ west of north,
laila [671]

Answer:

1.45544 J

Explanation:

See attachment

5 0
3 years ago
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