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hichkok12 [17]
2 years ago
11

Please solve this worksheet for me It’s very important please don’t write unnecessary things or else I’ll report please help me

i request

Physics
1 answer:
Sonja [21]2 years ago
3 0

Answer:

1. 231.25 mL

2. 218.75 kPa

3. 6.21 L

4. 427.5 mL

5. 83.01 mL

6. 307.6 mL

7. 1.15 L

8. 440.4 mL

Explanation:

This questions describes Boyle's law. Using Boyle's law equation to solve the questions as follows:

P1V1 = P2V2

Where;

P1 = initial pressure

P2 = final pressure

V1 = initial volume

V2 = final volume

Please find the answers to each question as an attachment.

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chubhunter [2.5K]

I learned the equation as P•V=k•T .

So x=1, y=1, and z= -1 .

3 0
2 years ago
A cyclist maintains a constant velocity of 6.7 m/s headed away from point
JulsSmile [24]
His position x = V* t = 6.7*56= 375.2 m from point a
3 0
3 years ago
A proton moving at 6.60 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.60 10-1
Hoochie [10]

Hi there!

We can use the following equation for a point charge in a magnetic field:


\large\boxed{F_B = qv \times B}

F_B = Force due to magnetic field (7.6 × 10⁻¹³N)
q = Charge of particle (1.6 × 10⁻¹⁹ C)

v = velocity of particle (6.6 × 10⁶ m/s)

B = Magnetic field strength (1.8 T)

Or, without the cross product:
F_B = qvBsin\theta

θ = angle between particle's velocity and field

We can rearrange to solve for theta:
\frac{F_B}{qvB} = sin\theta\\\\\theta = sin^{-1} (\frac{F_B}{qvB})

Solve for theta:
\theta = sin^{-1} (\frac{7.6*10^{-13}}{(1.6*10^{-19})(6.6*10^6)(1.80)}) = \boxed{23.57^o}

4 0
2 years ago
A thin rod of length 0.83 m and mass 110 g is suspended freely from one end. It is pulled to one side and then allowed to swing
VARVARA [1.3K]

Explanation:

(a)  The given data is as follows.

    Length of the rod, L = 0.83 m

    Mass of the rod, m = 110 g = 0.11   (as 1 kg = 1000 g)

 At the lowest point, angular speed of the rod (\omega) = 5.71 rad/s

First, we will calculate the rotational inertia of the rod about an axis passing through its fixed end is as follows.

      I = I_{CM} + mh^{2}

        = \frac{1}{12}mL^{2} + m(\frac{L}{2})

        = \frac{1}{12} \times 0.11 \times (0.83)^{2} + 0.11 \times \frac{0.83}{2}

        = 0.00631 + 0.415

        = 0.42131 kg m^{2}

Therefore, kinetic energy of the rod at its lowest point will be calculated as follows.

             K = \frac{1}{2}I \omega^{2}

                = \frac{1}{2} \times 0.42131 kg m^{2} \times (5.71)^{2}

                = 6.86 J

Hence, kinetic energy of the rod at its lowest point is 6.86 J.

(b)   According to the conservation of total mechanical energy of the rod, we have

         K_{i} + U_{i} = K_{f} + U_{f}

           K_{i} = U_{f} - U_{i}

or,      mgh = K = 6.86 J

Therefore,      h = \frac{6.86}{mg}

                          = \frac{0.63}{0.11 \times 9.8}

                          = 0.584 m

Hence, the center of mass rises 0.584 m far above that position.

6 0
3 years ago
Read 2 more answers
Three children are riding on the edge of a merry‑go‑round that has a mass of 10^5 and a radius of 1.80 m . The merry‑go‑round is
zimovet [89]

Answer: 22.01 rpm.

Explanation:  

If we assume no external torques are present, the angular momentum must be conserved.

L1 = L2

I1 . ω1 = I2 ω2 (1)

I1 = MR2 /2 + m1. R2 + m2. R2 + m3. R2  

I2 = MR2 /2 + m1. R2 + m3. R2

(as the 28.0 kg child moves to the center, he has no part anymore in the rotational inertia)

As I units are SI units, it is advisable to convert the angular velocity to rad/sec, as follows:

22.0 rev/min. (2π/rev) . (1min/60 sec) = 2.3 rad/sec

Solving for ω2 in (1):

ω2 = 2.31 rad/sec = 2.31 (1 rev/2 π). (60sec/1min) = 22.01 rpm

3 0
2 years ago
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