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Korvikt [17]
3 years ago
15

Two men decide to use their cars to pull a truck stuck in mud. They attach ropes and one pulls with a force of 598 N at an angle

of 29◦ with respect to the direction in which the truck is headed, while the other car pulls with a force of 941 N at an angle of 23◦ with respect to the same direction. 598 N 29 ◦ 941 N 23 ◦ What is the net forward force exerted on the truck in the direction it is headed? Answer in units of N
Physics
1 answer:
kondaur [170]3 years ago
5 0

Answer:

magnitude of the force is

F = 1537 N

direction of the force is given as

\theta = 25.3 degree

Explanation:

As we know that the first force is 598 N at 29 degree

so we have

F_1 = 598(cos29 \hat i + sin29 \hat j)

F_1 = 523 \hat i + 290 \hat j

Now another force is 941 N at 23 degree

F_2 = 941(cos23\hat i + sin23\hat j)

F_2 = 866.2 \hat i + 367.7 \hat j

so we will have

F = F_1 + F_2

F = (523 + 866.2)\hat i + (290 + 367.7)\hat j

F = 1389.2\hat i + 657.7 \hat j

magnitude of the force is

F = \sqrt{1389.2^2 + 657.7^2}

F = 1537 N

direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{657.7}{1389.2}

\theta = 25.3 degree

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A particle on a spring moves in simple harmonic motion along the x axis between turning points at x1 = 95 cm and x2 = 135 cm. (i
uranmaximum [27]

Answer:

(i) x = 115\,cm, (ii) x = 95\,cm, (iii) x = 95\,cm

Explanation:

(i) x_{1} and x_{2} represent the points where particle has a velocity of zero and spring reach maximum deformation, Given the absence of non-conservative force and by the Principle of Energy Conservation, the position where particle is at maximum speed is average of both extreme positions:

x = 115\,cm

(ii) Maximum accelerations is reached at x_{1} and x_{2}.

x = 95\,cm

(iii) Greatest net forces exerted on the particle are reached at  x_{1} and x_{2}.

x = 95\,cm

8 0
3 years ago
) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffi
natali 33 [55]

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

4 0
3 years ago
MathPhys i need your help please helpppo
Lena [83]

Answer:

6.77 m/s

Explanation:

First, in the x direction:

Given:

Δx = 3.17 m

v₀ = v cos 30.8° = 0.859 v

a = 0 m/s²

Δx = v₀ t + ½ at²

(3.17 m) = (0.859 v) t + ½ (0 m/s²) t²

3.17 = 0.859 v t

3.69 = v t

Next, in the y direction:

Given:

Δy = 0.432 m

v₀ = v sin 30.8° = 0.512 v

a = -9.81 m/s²

Δy = v₀ t + ½ at²

(0.432 m) = (0.512 v) t + ½ (-9.81 m/s²) t²

0.432 = 0.512 v t − 4.905 t²

Two equations, two variables.  Solve for t in the first equation and substitute into the second equation:

t = 3.69 / v

0.432 = 0.512 v (3.69 / v) − 4.905 (3.69 / v)²

0.432 = 1.89 − 66.8 / v²

66.8 / v² = 1.458

v² = 45.8

v = 6.77

7 0
3 years ago
Net force help please.
nekit [7.7K]
The net force is the total force. Add 4 and 2 together and you get 6. Since 5 N are pushing against it, you subtract that from 6. The net force is 1 N.
6 0
3 years ago
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Eduardwww [97]

Answer: Increase in competition for abiotic factors.

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