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stellarik [79]
3 years ago
10

Gizmo Warm-up: Lifting a piano A pulley is a simple machine that is used to lift heavy objects. A pulley is a wheel with a groov

e for a rope. Pulling down on one end of the rope causes the other end to pull up. In the Object to lift menu choose Piano. What is the weight of the piano
Physics
1 answer:
Nat2105 [25]3 years ago
5 0

Answer:

bruuuuuuuuh

Explanation:

id                  

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determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund
wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is a_G= 6.78 m/s^2

<u>Explanation</u>:

N_A+N_B-Mg = 0

-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0

0.8N_B +0.8N_A = 975a_G

N_A+N_B = 9564.75 -------------(1)

-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0

-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0

-2.26N_A -0.06N_B= 0 ----------------(2)

Solving the equation (1) and(2)

N_A + N_B = 9564.75

-2.26N_A-0.06N_B=0

N_A = -260.85N

N_B = 9825.60N

\mu_s N_B + \mu_s N_A = 975a_G

0.8(9825.60)+0.8(-260.85) = 975a_Ga_G=\frac{7651.8}{975}a_G_1=7.4848m/s^2

Next lets assume that the front wheels contact with the ground N_A = 0

F_B = Ma_G

N_B = M_g

N_B - M_g = 0

N_B(b-a) –F_Bh = 0

F_B = 975a_G

N_B-975(9.8) = 0

N_B=9564.75N

9564.75(2.20 -1.82) -F_B(0.55)=0

\frac{3634.605}{0.55}=F_B

F_B = 6608.3

F_B = Ma_G

6608.3 = 975a_G

a_G = 6.7778 m/s^2

a_G_2 = 6.78m/s^2

Choosing the critical case

a_G = min(a_G_1 ,a_G_2)

a_G = min(7.848, 6.78)

a_G= 6.78 m/s^2

3 0
4 years ago
Which option correctly matches the chemical formula of a compound with its name?
77julia77 [94]

Answer:it’s A

Explanation:

4 0
3 years ago
Read 2 more answers
if the 50 kg object slows down to a velocity of 1 m/s how much kinetic energy does it have 100j 50j 25j or none
rodikova [14]

ek =  \frac{1}{2}m {v}^{2}

ek =  \frac{1}{2}(50) {(1)}^{2}

ek =  \frac{1}{2}(50)

ek = 25j

//

I'm not really sure but I do know that it's not 0 because the object is still moving, even if it's only moving at 1 m/s.

6 0
3 years ago
1. Molly carries a 30 kg bag of feed up a ramp which has a base of 4 m and a height of 2 meters. How
a_sh-v [17]

Answer:

it is b. 588 Nm

Explanation:

because am smart hehe...

6 0
3 years ago
Three charges, qA is +6.0 μC, qB is –5.0 μC, and qc is +6.0 μC, are located at the corners of a square with each side length at
lesya [120]

The net electric field at point D is determined as 3.95 x 10⁷ N/C.

<h3>Electric field at D due to charge A</h3>

E = kq/r²

where;

  • r is the distance between A and D
  • q is charge A

E(AD) = (9 x 10⁹ x 6 x 10⁻⁶)/(0.05²)

E(AD) = 2.16 x 10⁷ i N/C

<h3>Electric field at D due to charge B</h3>

E = kq/r²

where;

  • r is the distance between A and B
  • q is charge B

r² = 5²  + 5²

r² = 50

r = √50

r = 7.07 cm

E(BD) = (9 x 10⁹ x 5 x 10⁻⁶)/(0.0707²)

E(BD) = 9 x 10⁶ N/C

in x - direction = 9 x 10⁶ N/C x cos(45) = 6.36 x 10⁶ i  N/C

in y - direction = 9 x 10⁶ N/C x sin(45) = 6.36 x 10⁶j  N/C

<h3>Electric field at D due to charge C</h3>

E = kq/r²

where;

  • r is the distance between C and D
  • q is charge C

E(CD) = (9 x 10⁹ x 6 x 10⁻⁶)/(0.05²)

E(CD) = 2.16 x 10⁷ j N/C

<h3>Net electric field in x direction</h3>

Ei = 2.16 x 10⁷ i N/C  + 6.36 x 10⁶ i  N/C

Ei = 2.796 x 10⁷ i N/C

<h3>Net electric field in y direction</h3>

Ej = 2.16 x 10⁷ j N/C  + 6.36 x 10⁶j  N/C

Ej = 2.796 x 10⁷ j N/C

<h3>Resultant electric field at D</h3>

E = √Ei² + Ej²

E = √[(2.796 x 10⁷)² + (2.796 x 10⁷)²]

E = 3.95 x 10⁷ N/C

Thus, the net electric field at point D is determined as 3.95 x 10⁷ N/C.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

7 0
2 years ago
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