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tatuchka [14]
3 years ago
5

A 5.3 kg cat and a 2.5 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw. how far to the left of the pivot must

a 3.7 kg cat stand to keep the seesaw balanced?
Physics
1 answer:
nika2105 [10]3 years ago
7 0
Moment about the pivot must be equal for the seesaw to balance. Initially, the first cat and the bowl are at 2 m from the pivot.

The moment due to cat = 5.3*2 = 10.6 kg.m
The moment due to bowl = 2.5*2 = 5 kg.m
The unbalanced moment = 10.6 - 5 = 5.6 kg.m

Therefore, the 3.7 kg cat should stand at a distance x from the pivot in left to balance the 5.6 kg.m.
That is,
3.7*x = 5.6 => x = 5.6/3.7 = 1.5134 m to the left (on the side of the bowl)
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(3mrks) physics solar panel question
shtirl [24]
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6 0
3 years ago
The electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the
elixir [45]

Answer:

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Explanation:

When connected in series, the equivalence resistance, R(eq) is given as

R(eq) = (R₁ + R₂)

When connected in parallel, the equivalence resistance, R(eq) is given as

[1/R(eq)] = [(1/R₁) + (1/R₂)]

R(eq) = (R₁R₂)/(R₁ + R₂)

The parallel and series combination are connected to a battery of emf 39.0 V with negligible internal resistance. And the power supplied is measured.

But power supplied is given as

P = IV = (V/R) V = (V²/R)

When connected in series, the power supplied is given as

P = 48.0 W,

V = 39.0 V,

R = R(eq) = (R₁ + R₂)

48 = (39²/R)

R = (39²/48)

R = 31.6875 ohms

R = (R₁ + R₂) = 31.6875

(R₁ + R₂) = 31.6875 (eqn 1)

When connected in series, the power supplied is given as

P = 256.0 W,

V = 39.0 V,

R = R(eq) = (R₁R₂)/(R₁ + R₂)

256 = (39²/R)

R = (39²/256)

R = 5.9414 ohms

R = R(eq) = (R₁R₂)/(R₁ + R₂) = 5.9414

(R₁R₂)/(R₁ + R₂) = 5.9414

But, recall eqn 1

(R₁ + R₂) = 31.6875

(R₁R₂)/(R₁ + R₂) = 5.9414

Substituting for (R₁ + R₂)

(R₁R₂)/(R₁ + R₂) = (R₁R₂)/31.6875 = 5.9414

(R₁R₂) = 31.6875 × 5.9414 = 188.2683

R₁ = (188.2683/R₂)

(R₁ + R₂) = 31.6875

Substituting for R₁

(188.2683/R₂) + R₂ = 31.6875

multiply through by R₂

188.2683 + R₂² = 31.6875R₂

R₂² - 31.6875R₂ + 188.2683 = 0

Solving the quadratic equation

R₂ = 23.77 ohms or 7.92 ohms

If R₂ = 23.77 ohms, R₁ = 7.92 ohms

If R₂ = 7.92 ohms, R₁ = 23.77 ohms

Since the question explains that R₁ > R₂

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Hope this Helps!!!

3 0
3 years ago
Can someone help me ?
hammer [34]

Answer:

1)    Time interval                 Blue Car                      Red Car

          0 - 2 s                Constant Velocity           Increasing Velocity

          2 - 3 s                Constant Velocity           Constant Velocity

          3 - 5 s                Constant Velocity           Increasing Velocity

          5 - 6 s                Constant Velocity           Decreasing Velocity

2) For Red and Blue car y₂  = 120       v = \frac{y_{2}-y_{1}}{t_{2}-t_{1}} = \frac{120-0}{6-0} = 20 m/s

     We get the same velocity for two cars because it is the average velocity of the car at the given interval of time. It is measured for initial and final position.

3)   At t = 2s, the cars are the same position, and are moving at the same rate

                    Position - same

                    Velocity - same

The position-time graph shares the same spot for two cars.

4 0
3 years ago
The angle between incident ray and reflected ray is 130.what is the value of angle of incidence​
Artyom0805 [142]

Answer:

65

Explanation:

as i = r , so i + i = 130

so , i = 130/2 =65

7 0
3 years ago
Read 2 more answers
What is the momentum of a 0.15 kilgram baseball moving at 20 m/s?
natita [175]
Momentum = mv
= .15 *20
= 3 kgm/s^2
5 0
3 years ago
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