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sweet [91]
3 years ago
10

Two people carry a heavy electric motor by placing it on a light board 1.80 m long. One person lifts at one end with a force of

350 N, and the other lifts the opposite end with a force of 620 N.
What is the weight of the motor?
Physics
1 answer:
bixtya [17]3 years ago
6 0

Answer:

W=970\ N

Explanation:

When two people carry a weight placing it on a very light board then we neglect the weight of the board and the board acts like a beam with two mass placed over it and two end point supports are here provided by men.

Then balancing the forces in the vertical direction, weight of the motor is:

W=350+620

W=970\ N

These men are able to carry the motor in equilibrium condition because they have placed the motor (considering it a point mass) at a position such that the moment due to the lifting forces are equal and opposite.

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erastovalidia [21]

Answer:

KE = 1/2mv^2

KE = 1/2(24)(3^2)

KE = 12(9)

KE = 108 J

Let me know if this helps!

5 0
3 years ago
What are two reasons why a home computer scanner requires
alexandr1967 [171]

Answer: C and D

Explanation: a p e x

4 0
2 years ago
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A vector makes an angle of 30° with the positive x-axis.
kkurt [141]

Answer:

W = 112.58 N-unit

Explanation:

Given:

- Force F = 10 N

- Angle Q of force with x axis: 30 degrees

- distance to be moved d = 13 units along + x axis

Find:

Work Done by the force F:

Solution:

The work by force in positive x direction can only be done if the both the direction of distance traveled and direction of force are parallel vectors. Hence we compute the component of Force F in x direction F_x:

                                       F_x = F*cos(Q)

                                       F_x = 10*cos(30)

                                       F_x = 8.66 N

        Hence,

                                  Work Done by force

                                        W = F_x * d

                                       W = 8.66 * 13

                                  W = 112.58 N-unit

5 0
3 years ago
A 0.877-kg ball has an elastic, head-on collision with a second ball that is initially at rest. The second ball moves off with h
Art [367]

Answer: the mass of the second ball is 2.631 kg

Explanation:

Given that;

m1 = 0.877 kg

Initial velocity = V0

Initial momentum = m1 × V0

final velocity of m1 is u1, final velocity of m2 is u2 = v0/2

now final momentum = m1 × u1 + m2 × u2

using momentum conservation;

m1×V0 = m1×u1 + m2×v0/2

m1×(v0 - u1) = m2×V0/2 ----- let this be equation 1

Now, for elastic collision;

m1×v0²/2 = m1×u1²/2 + m2×(v0/2)²/2

m1×(v0² - u1²) = m2×(v0/2)² --------- let this be equation 2

now; equation 2 / equation 1

: V0 + u1 = v0/2

2V0 + 2u1 = V0

2u1 = V0 - 2V0

u1 = -V0/2

now we insert in equ 1

 m1×3V0/2= m2×V0/2

m1 × 3 = m2

m2 = 0.877 × 3

m2 = 2.631 kg

Therefore, the mass of the second ball is 2.631 kg

7 0
3 years ago
PLEASE ASAP<br><br> Giving a brainlist
-Dominant- [34]
Between g and h the object is not moving.
4 0
3 years ago
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