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OLEGan [10]
3 years ago
9

A 1.5 kg copper block is given an initial speed of3.00m/s on a

Physics
1 answer:
ankoles [38]3 years ago
6 0

Answer:

a. 0.01 C

b. dissipated to outside environment

Explanation:

Let the specific heat of copper be 0.3846 kJ/kg-K or 384.6 J/kg-C

(a)The original kinetic energy of the block is:

E_k = \frac{mv^2}{2} = \frac{1.5*3^2}{2} = \frac{1.5*9}{2} = 6.75 J

As 85% of this kinetic energy is converted to block internal heat energy, with specific heat we can calculate the rise in temperature:

E_h = 0.85E_k = 0.85*6.75 = 5.7375 J

mc_c\Delta T = 5.7375

1.5*384.6\Delta T = 5.7375

\Delta T = \frac{5.7375}{1.5*384.6} \approx 0.01^oC

(b) the remaining 15% energy would probably be dissipated to outside environment as heat energy

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3 years ago
When a hailstone is at a height of 2.00km it’s mass is 2.50g what is it’s potential energy?
scZoUnD [109]

Answer:

EP = 49.05Joules (J)

Explanation:

The equation for Potential energy (EP) is

EP = m g h

We are given the values below (do convert them into SI units)

m = 0.0025kg

h = 2000m

g = 9.81m/s^{2}

Substitute the values into the equation and solve for EP

EP = 0.0025 * 2000 * 9.81

EP = 49.05Joules (J)

6 0
3 years ago
object is placed 15cm in front of a plane mirror. If the mirror is moved further 5cm away from the object and the image is​
solong [7]

Answer:

The image will most likely be 20cm in front the mirror since the mirror was placed further 5cm.

4 0
3 years ago
Read 2 more answers
Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s
alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

  • Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
  • Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
  • vₓ₀ = v * cos θ (1)
  • where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
  • Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

        x_{f} = v_{ox} * t = v_{o} * cos \theta * t  (2)

  • Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

        cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)

  • ⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

  • At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
  • Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
  • vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

  • At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

  • Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

       \Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)

  • Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
  • v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
  • Replacing (7) in (6), we get:

       \Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)

e)

  • When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
  • The horizontal component, since it keeps constant, is just v₀x:
  • v₀ₓ = 13.7 m/s
  • The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       v_{fy} = v_{oy} - g*t  (9)

  • Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s  (10)

  • Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

       v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)

3 0
3 years ago
Phil is riding a scooter and pushes off the ground with his foot. this causes him to accelerate at 12 m /s. Phil weighs 600 N. h
Dvinal [7]

Answer:

734.16 kg m/s^{2}

Explanation:

The problem is asking for the Force of pushing off the ground.

  • The formula of Force is: F = mass x acceleration

Given = <em>Mass</em>: 600 newtons (N)

             <em>Acceleration</em>: 12 m/s^{2}

We have to convert the mass into kg first. Remember that <u>1 kg is equal to 9.80665 newtons.</u>

Let x be the<em> mass in newtons</em>.

Let's convert: \frac{1 kg}{9.80665 N} x \frac{x}{600 N} = \frac{600}{9.80665} = 61.18 kg

Phil's weight is 61.18 kg

Let's go back to finding the force.

F = m x a

F = 61.18 kg x 12 m/s^{2}

F = 734.16 kg m/s^{2}

7 0
3 years ago
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