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Alchen [17]
3 years ago
11

Two ropes are attached to a 40-kg object. The first rope applies a force of 50 N and the second, 40 N. If the two ropes are perp

endicular to each other, what is the resultant magnitude of the acceleration of the object?
Physics
1 answer:
Montano1993 [528]3 years ago
4 0

To solve this problem we will use the concepts related to the resulting Vector Force product of two components, that is,

|\vec{F}| = \sqrt{F_x^2+F_y^2}

If we take the Force of 50 N as the force in the X direction and the Force of 40 N in the Y direction we will have to:

|\vec{F}| = \sqrt{F_x^2+F_y^2}

|\vec{F}| = \sqrt{(50)^2+(40)^2}

|\vec{F}| = 64.03N

Finally, since Newton's second law, acceleration can be determined as

F = ma

a = \frac{F}{m}

a = \frac{ 64.03}{40}

a = 1.6m/s^2

Therefore the resultant magnitude of the acceleration of the object is 1.6m/s^2

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7 0
3 years ago
Anyone know how to do this?
MrMuchimi

The voltage from one side of the battery all the way around to the other side of the battery is 12v .

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Just like it says in choice-C.

7 0
2 years ago
A ball is thrown vertically up from the edge of a cliff with a speed of 8 m/s, how high is
anygoal [31]

Explanation:

s=d/t

d=s×t

d=8×16.4

d=131.2

distance is 131.2m.

7 0
3 years ago
Riding in a car, you suddenly put on the brakes. As you experience it inside the car, do Newton's law apply? Do they apply as se
alisha [4.7K]

Answer with Explanation:

Newton's laws are applicable for inertial frames of reference which is a frame which is not accelerating when seen from the observer standing on earth.

For the person as he presses the brakes his frame is a decelerating frame of reference hence he cannot apply the newtons laws of motion as they are in their original form but if he analyses the motion he has to apply a correction known as  pseudo-force on the object he is analyzing. Pseudo Force has no basis in newton's laws but are a correction that needs to be applied if he wishes to analyse the motion from non inertial frame of reference

While as a person standing on earth outside the car since his frame is an inertial frame of reference he can apply newton's laws of motion without any correction.  

3 0
3 years ago
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
3 years ago
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