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2 years ago

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A 1500 kg car is pushing a 4000 kg truck. The car and truck are accelerating at 2.0 m/s^2. Assuming that the frictional force on

the truck is negligible, what force is the truck exerting on the car?

1 answer:

Crazy boy [7]2 years ago

7 0

To solve this problem we will use the Force equation according to the definition given in Newton's second law. There we have that the Force is equal to

$F =ma$

Where,

m = mass

a = acceleration

Our values are given as

$m_1 = 1500 kg$

$m_2 = 4000 kg$

$a = 2.0 m/s^2$

Considering that both mass are equal to one, we have that:

$F = (m_1+m_2)* a$

$F = (1500 + 4000)(2.0)$

$F = 11000 N = 11kN$

Therefore the truck exert a force on the car of 11kN

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andrew11 [14]

Answer:

Explanation:

The direction of a magnetic field indicates where the magnetic inluence on the electric charges are directed to.

From the given question, we are to determine the direction of the magnetic field bnet at a point A.

Also, having the notion that the currents in the two wires have equal magnitudes, Then:

$\bar{B_{net}} = \bar{B_1} + \bar{B_2}$

$\bar{B_{net}} = \frac{\mu_oI}{2 \pi r } \bar {k}+ \frac{\mu_oI}{2 \pi r } \bar {k}$

$\bar{B_{net}} = \frac{2 \mu_oI}{2 \pi r } \bar {k} \ out$

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2 years ago

A water-skier with weight Fg = mg moves to the right with acceleration a. A horizontal tension force T is exerted on the skier b

Degger [83]

Answer:

The correct relationships are T-fg=ma and L-fg=0.

(A) and (C) is correct option.

Explanation:

Given that,

Weight Fg = mg

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Drag force = Fa

Vertical force = L

We need to find the correct relationships

Using balance equation

In horizontally,

The acceleration is a

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In vertically,

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$w=L$

$mg-L=0$

Put the value of mg

$L-fg=0$ ....(II)

Hence, The correct relationships are T-fg=ma and L-fg=0.

(A) and (C) is correct option.

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RSB [31]

Answer:

r = 4.44 m

Explanation:

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B = ρ g V

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B - W = 0

B = W

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σ = W / A

W = σ A

The area of a sphere is

A = 4π r²

W = W₁ + σ 4π r²

The volume of a sphere is

V = 4/3 π r³

Let's replace

ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

P V = n RT

P = ρ RT

ρ = P / RT

P / RT g 4/3 π r³ - σ 4 π r² = W₁

r² 4π (P/3RT r - σ) = W₁

Let's replace the values

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As the independent term is very small we can despise it, to find the solution

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2 years ago

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Artyom0805 [142]

Answer:

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

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F = m r w²

in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

Explanation:

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now we can say

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2 years ago