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belka [17]
3 years ago
9

7. An outfielder throws a baseball to the first baseman at a speed of 19.6

Physics
1 answer:
vlada-n [284]3 years ago
3 0
Problem the mount I cough my so the newer I 75.4
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3 years ago
A 2 kg object has a specific heat capacity of 1,700 J/(kg \cdot⋅oC)
Nutka1998 [239]

The amount of heat needed to raise the temperature of a 2kg object from 15°C to 25°C is 34000J.

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

  • The amount of heat absorbed by an object can be calculated by using the following expression:

  • Q = m.c.∆T

Where;

  1. Q = amount of heat absorbed or released (J)
  2. m = mass of object
  3. c = specific heat capacity (J/g°C)
  4. ∆T = change in temperature (°C)

  • According to this question, 2 kg object has a specific heat capacity of 1,700J/kg°C and was raised from a temperature of 15 Celsius to 25 Celsius. The heat absorbed is calculated as follows:

  • Q = 2 × 1700 × {25 - 15}

  • Q = 3400 × 10

  • Q = 34000J

  • Therefore, the amount of heat needed to raise the temperature of a 2kg object from 15°C to 25°C is 34000J.

Learn more about how to calculate heat absorbed at: brainly.com/question/11194034?referrer=searchResults

8 0
2 years ago
A car accelerates from 20.0 m/s to 28.0 m/s over a distance of 50.0 m. What is the car’s acceleration?
Blababa [14]

Answer:

Explanation:

V^2=V^2_o+2a(x_f-x_i)

28^2=20^2+2a(50)\\784=400+100a\\384=100a\\a=3.84m/s^2

5 0
3 years ago
Freeeeeeeee poinnttttsssss
RUDIKE [14]

Answer:

k

Explanation:

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5 0
3 years ago
A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vert
WARRIOR [948]

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta

\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta

\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h

\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h

h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g} (1)

Where:

h - Maximum height of the wood block, in meters.

v - Initial speed of the block, in meters per second.

\mu_{k} - Kinetic coefficient of friction, no unit.

g - Gravitational acceleration, in meters per square second.

m - Mass, in kilograms.

s - Distance travelled by the wood block along the wooden ramp, in meters.

\theta - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that v = 10\,\frac{m}{s}, \mu_{k} = 0.20 and g = 9.807\,\frac{m}{s^{2}}, then the height reached by the block above its starting point is:

h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

h = 4.249\,m

The wood block reaches a height of 4.249 meters above its starting point.

5 0
3 years ago
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