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anygoal [31]
2 years ago
7

How much energy is needed to change the temperature of 50g of water 15°c​

Physics
1 answer:
zzz [600]2 years ago
7 0

Explanation:

This question is not feasible. There is no way to calculate the energy needed because the question is missing the final temperature

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A disk rotates at constant angular acceleration, from angular position θ1 = 16.0 rad to angular position θ2 = 76.0 rad in 5.30 s
Oliga [24]

Answer:

(a) the angular velocity at θ1 is 11.64 rad/s

(b) the angular acceleration is 0.12 rad/s^{2}

(c) the angular position was the disk initially at rest is - 428.27 rad

Explanation:

Given information :

θ1 = 16 rad

θ2 = 76 rad

ω2 = 11 rad/s

t = 5.3 s

(a) The angular velocity at θ1

First, we use the angular motion equation for constant acceleration

Δθ = (ω1+ω2)t/2

θ2 - θ1 = (ω1+ω2)t/2

ω1 + ω2 = 2 (θ2 - θ1) / t

ω1 = (2 (θ2 - θ1) / t ) - ω2

     = (2 (76-16) / 5.3) - 11

     = 11.64 rad/s

(b) the angular acceleration

ω2 = ω1 + α t

α t = ω2 - ω1

α = (ω2 - ω1)/t

  = (11.64 - 11) / 5.3

  = 0.12 rad/s^{2}

(c) the angular position was the disk initially at rest, θ0

at rest ω0 = 0

ω2^2 = ω01  t + 2 α Δθ

2 α Δθ = ω2^2

θ2 - θ0 = ω2^2  /  2 α

θ0 = θ2 -  (ω2^2) / 2 α

  = 76 - (11^{2}/ 2 x 0.12

  = 76 - 504.16

  = - 428.27 rad

4 0
3 years ago
In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of t
muminat
<h2><em><u>⇒</u></em>Answer:</h2>

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Step-by-Step Solution:

Solution 35PE

This question discusses about the increased range. So, we shall assume that the angle of jumping will be  as the horizontal range is maximum at this angle.

Step 1 of 3<

/p>

The legs have an extension of 0.600 m in the crouch position.

So,  m

The person is at rest initially, so the initial velocity will be zero.

The acceleration is  m/s2

Acceleration  m/s2

Let the final velocity be .

Step 2 of 3<

/p>

Substitute the above given values in the kinematic equation  ,

m/s

Therefore, the final velocity or jumping speed is  m/s

Explanation:

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The resistance at operating temperature is R = V/I = 2.9 V / 0.23A = 12.61 ohmsT from R – R0 = Roalpha (T – T0), we find that:T = T0 + 1/alpha (R/R0 -1) = 20 degrees Celsius + (1/ 4.3 x 10^-3/K) (12.61 ohms/ 1.1 ohms – 1)T = 2453.40 degrees Celsius
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Option C Magnetic force
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