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Nadusha1986 [10]
2 years ago
10

Calculate the magnitude and the direction of the resultant forces​

Physics
1 answer:
snow_tiger [21]2 years ago
6 0

answer:

resultant = 127.65 in the positive direction

explanation:

F1 = 50N , F2 = 40N, f3 = 55N , f4 = 60N

Fy = 50 sin 50 = 50 × -0.26 = -13

Fx = 40 cos 0 = 40×1 = 40

fx = 55 cos 25 = 55×0.99 = 54.45

Fy = 60 sin 70 = 60 × 0.77 = 46.2

resultant = -13+40+54.45+46.2 = 127.65 in the positive direction

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Consider the following objects and their locations. Object a 2-kg object 5 cm above the floor Object b 2-kg object 120 cm above
Nataly [62]

Answer:

The gravitational acceleration is same for all objects.

a = b = c = d

Explanation:

Acceleration due to gravity or gravitational acceleration is the force exerted by Earth on unit mass of an object.

Acceleration due to gravity doesn't depend on the height of the object when the height is object is near to the surface of the Earth. Only when the height is comparable to the radius of the Earth, the value of gravitational acceleration changes.

But for the objects here, the gravitational acceleration is independent of the mass or height of the objects and has a constant value of 9.8 m/s².

Therefore, the gravitational acceleration of all the objects is same.

If 'a', 'b', 'c', and 'd' represent gravitational accelerations of objects 'a', 'b', 'c', and 'd' respectively, then a = b = c = d.

5 0
2 years ago
A 10 cm long tensile specimen with a yield stress of 350 MPa is pull to a stress of 300 MPa and then the load is released. What
Advocard [28]

Answer:

The length of the specimen after the load is released is 11.67 cm

Explanation:

Given;

yield stress, Y = 350 MPa

ultimate tensile stress, T = 300 MPa

Elongation factor, e = yield stress, Y / ultimate tensile stress, T

Elongation factor, e = 350 Mpa / 300 Mpa

Elongation factor, e = 1.1667

New length of the specimen = 1.1667 x 10 cm = 11.67 cm

Therefore, when the load is released from 10 cm long tensile specimen, the length of the specimen becomes 11.67 cm

3 0
3 years ago
A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory?
Korolek [52]

Answer:

Net external force acting on the rock when it is at the top of its trajectory is force due to gravity (mg).

Explanation:

The forces acting on a rock thrown up are force due to gravity and air resistance. Air resistance is directly proportional to velocity of rock, when velocity is zero air resistance is zero. When it is at the top of its trajectory its velocity is zero. So air resistance is also zero. Hence only gravitational force acts on the rock.

Net external force acting on the rock when it is at the top of its trajectory is force due to gravity (mg).

6 0
3 years ago
Help me pleaseeeeeeddd​
Vladimir79 [104]

Answer:

Explanation:

a )

According to graph the object moves with constant velocity of 10 m /s during first 8 s and then its velocity changes to - 5 m /s from 8 th second upto 12 th second .

Initial displacement = 8 m

displacement during 8 s = velocity x time

= 8 x 10 = 80 m

total displacement = 88 m

b )

displacement during period from 8 th to 12 th s

= - 5 x 4 = - 20 m

total displacement after 12 s

= 88 - 20 = 68 m

c ) average speed = total distance covered during 12 s / total time

= (80 + 20) / 12

= 8.33 m /s

average velocity = total displacement during 12 s  / total time

= 80 - 20 / 12  

= 60 / 12

= 5 m /s .

7 0
2 years ago
We can determine the ____________ of a wave when given the frequency and the wavelength.
kotykmax [81]
<span>We can determine the "Velocity" of</span><span> a wave when given the frequency and the wavelength.

So, option B is your answer.

Hope this helps!
</span>
6 0
3 years ago
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